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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Algebra 1
Module 4
Edited for use by the North Thurston Public Schools
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 1: Multiplying Polynomial Expressions
Example 1 The two rectangles below each represent a multiplication of two terms. Specifically, the unshaded
rectangle shows 2π₯(π₯) and the shaded rectangle shows 2π₯(8).
The area of the unshaded rectangle is 2π₯2 and the area of the shaded rectangle is 16π₯. The total area
of the two is then 2π₯2 + 16π₯
If we join the two rectangles, the total area remains the same, but the product represented is a little
different.
This same idea was covered back in Module 1, Topic B. We are revisiting it here because now we want
to be able to work this process backwards --- if we know that the result is 2π₯2 + 16π₯, can we find what
was multiplied to create it?
But in order to be good at that, we need to be very good at both monomial and binomial multiplication.
1. Copy and complete the following monomial multiplications in your notes:
a. 4π₯(3π₯ + 5) b. 5π₯(2π₯ + 3)
π₯
2π₯ 2π₯
8
By the end of this lesson, you should be able to:
multiply a monomial and a binomial
multiply two binomials
identify special products of binomials
π₯
2π₯
8
This now represents 2π₯(π₯ + 8)
which means
2π₯(π₯ + 8) is the same as 2π₯2 + 16π₯
Page 1
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 2
Using a Table as an Aid
The model used in Example 1 is known as an area model, or a generic rectangle. It is best for problems
that involve only addition, since each expression is supposed represent a measurement of length.
Without a context such as length, we canβt be sure that an algebraic expression represents a positive
quantity. But we can still mimic a generic rectangle by changing it into a table. The table serves to
remind us of the area model even though it does not represent area.
For example, the table below shows the product of (3π₯ β 2)(π₯ β 5). Each cell in the table represents
the multiplication of the term at the top of the column with the term at the beginning of the row:
π₯ β5
3π₯ 3π₯2 β15π₯
β2 β2π₯ +10
This table helps us to organize the multiplication of (3π₯ β 2)(π₯ β 5) and see the resulting terms. The
answer to the multiplication is: 3π₯2 β 15π₯ β 2π₯ + 10 which simplifies to: 3π₯2 β 17π₯ + 10.
Without the Aid of a Table
Regardless of whether or not we use a table as an aid, multiplying of two binomials is an application of
the distributive property. Both terms of the first binomial distribute over the second. In other words:
(3π₯ β 2)(π₯ β 5) becomes 3π₯(π₯ β 5) β 2(π₯ β 5)
Since there are 2 terms in each binomial, there are 4 multiplications that must be done. But itβs
possible to do these multiplications without separating the first binomial. The process is often
referred to by the acronym βFOILβ. This stands for First, Outside, Inside, and Last.
Here is what the process looks like. The arrows match each step of the distribution with the resulting
partial product.
(3π₯ β 2)(π₯ β 5)
3π₯2
β15π₯
β2π₯
+10 }
3π₯2 β 15π₯ β 2π₯ + 10
Again, we would combine the like terms in the middle to get a final answer of:3π₯2 β 17π₯ + 10
first
last
inside
outside
Page 2
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
We leave the answer in standard order, meaning the terms are in order from highest power to lowest.
2. Copy and complete the following binomial multiplications in your notes. Leave your answer in
standard order:
a. (4π₯ β 5)(π₯ + 2) b. (2π₯ + 7)(3π₯ β 1)
Example 2 Certain types of binomial multiplication are known as βspecial productsβ because their multiplication
results always follow the same pattern. Consider these examples:
(π₯ + 5)(π₯ β 5) becomes
π₯2 β 5π₯ + 5π₯ β 25 which becomes
π₯2 β 25
(π₯ + 7)(π₯ β 7) becomes
π₯2 β 7π₯ + 7π₯ β 49 which becomes
π₯2 β 49
(3π₯ + 5)(3π₯ β 5) becomes
9π₯2 β 15π₯ + 15π₯ β 25 which becomes
9π₯2 β 25
(4π₯ + 7)(4π₯ β 7) becomes
16π₯2 β 28π₯ + 28π₯ β 49 which becomes
16π₯2 β 49
This pattern is usually referred to as βthe difference of two squaresβ because the final answer is
always a subtraction (thatβs the βdifferenceβ part of the name) of two terms that are perfect squares.
This happens whenever the two binomials are conjugate pairs, meaning that they have the same
numbers, but with an opposite sign in between them. For example:
These are conjugate pairs: These are NOT conjugate pairs:
(3π₯ + 5)(3π₯ β 5)
(4π₯ + 7)(4π₯ β 7)
(π₯ + 3)(π₯ β 3)
(3π₯ + 5)(π₯ β 5)
(4π₯ + 7)(4π₯ + 7)
(π₯ + 3)(2π₯ β 3)
When we have conjugate pairs, we can skip over the middle step of the multiplication and go directly
to a difference (subtraction) of 2 squares, like this:
(π₯ β 9)(π₯ + 9)
becomes
π₯2 β 81
(π₯ + 11)(π₯ β 11) becomes
π₯2 β 121
(5π₯ + 6)(5π₯ β 6) becomes
25π₯2 β 36
(3π₯ β 10)(3π₯ + 10) becomes
9π₯2 β 100
Page 3
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
3. Copy the following problems into your notes. Use the difference of two squares shortcut to
find the product of the binomials.
a. (7π₯ + 1)(7π₯ β 1) b. (8π₯ β 3)(8π₯ + 3)
Example 3 The other βspecial productβ is created when the multiplication involves the same binomial twice,
meaning that the binomial is being squared.
Consider these examples:
(π₯ + 5)(π₯ + 5) becomes
π₯2 + 5π₯ + 5π₯ β 25 which becomes
π₯2 + 10π₯ + 25
(π₯ β 7)(π₯ β 7) becomes
π₯2 β 7π₯ β 7π₯ + 49 which becomes
π₯2 β 14π₯ + 49
(3π₯ + 5)(3π₯ + 5) becomes
9π₯2 + 15π₯ + 15π₯ β 25 which becomes
9π₯2 + 30π₯ + 25
(4π₯ β 7)(4π₯ β 7) becomes
16π₯2 β 28π₯ β 28π₯ β 49 which becomes
16π₯2 β 56π₯ + 49
You can see that the individual products are extremely similar between this and the previous example.
The only change is to the sign of one of the two middle terms. That sign change is significant because
now, instead of the two middle terms cancelling each other out, they βdouble upβ because they are both
the same.
So the pattern here is that the binomial, squared, becomes a trinomial where the first and third terms
are perfect squares, and the middle term is twice as big as the product of the two numbers:
(3π₯ + 5)(3π₯ + 5) πππππππ 9π₯2 + 30π₯ + 25
Often, the problem is presented as a binomial squared, rather than as two identical binomials:
(4π₯ β 7)2 πππππππ 16π₯2 β 56π₯ + 49
Notice that the third term of the result is positive. Thatβs because (β7)2 means multiplying β7 Γ β7,
and two negatives multiply to create a positive result.
3π₯, squared 5 squared 3π₯ times 5 times 2
4π₯, squared β7, squared 4π₯ times β7 times 2
Page 4
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1
ALGEBRA I
Lesson 1: The Distributive Property Last edited on 2/27/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
CAUTION: if you use a calculator to evaluate (β7)2, be sure to put parentheses around the β7.
Otherwise, the calculator will take 7, square it, and then make it negative, which is not the same.
4. Copy the following problems into your notes. Use the binomial squared shortcut to find the
result of the multiplication.
a. (π₯ + 6)2 b. (2π₯ β 9)2
Problem Set
Complete the following multiplications. Use shortcuts when possible.
1. (π₯ β 10)2 2. (5π₯ + 8)2
3. (π₯ β 8)(π₯ + 8) 4. (2π₯ β 3)(2π₯ + 3)
5. (3π₯ β 7)2 6. (π₯ β 1)2
7. (2π₯ β 9)(2π₯ + 3) 8. (π₯ + 5)(4π₯ + 5)
9. (π₯ β 10)(π₯ + 2) 10. (π₯ + 8)(π₯ + 3)
11. (π₯ + 11)(π₯ β 5) 12. (π₯ β 7)(π₯ β 4)
13. (3π₯ β 5)(2π₯ + 7) 14. (9π₯ β 1)(π₯ + 3)
15. (4π₯ β 5)(π₯ β 2) 16. (π₯ + 11)(2π₯ β 7)
17. (6π₯ + 5)(π₯ + 2) 18. (4π₯ + 9)(2π₯ β 3)
Page 5
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 2: Multiplying and Factoring Polynomial Expressions
Before we begin, letβs review some vocabulary: FACTOR: a term involved in a multiplication Example: 4 is a factor of 12 π₯π₯ is a factor of 12π₯π₯ 4π₯π₯ is a factor of 12π₯π₯ (π₯π₯ + 4) is a factor of (π₯π₯ + 4)(π₯π₯ + 3) POLYNOMIAL IN STANDARD ORDER: an algebraic expression with more than one term where the terms are in order from highest power to lowest power Example: 4π₯π₯2 + 12π₯π₯ β 8 LEADING COEFFICIENT: when a polynomial is in standard order, the leading coefficient is the coefficient of the first term (the term with the highest power) Example: the leading coefficient of 4π₯π₯2 + 12π₯π₯ β 8 is 4 TRINOMIAL: a polynomial with 3 terms, like 4π₯π₯2 + 12π₯π₯ β 8 BINOMIAL: a polynomial with 2 terms, like 4π₯π₯ + 12 Example 1 In the last lesson, we reviewed how to multiply binomials such as (π₯π₯ + 7)(π₯π₯ + 3), using either a table or the FOIL method. The multiplication would look like this:
or And in both methods, the final answer is: π₯π₯2 + 10π₯π₯ + 21.
π₯π₯ + 7
π₯π₯
+
3
π₯π₯2 7π₯π₯
3π₯π₯ 21
By the end of this lesson, you should be able to: β’ multiply two binomials β’ when given a trinomial whose leading coefficient is 1, factor into two binomials
(π₯π₯ + 7)(π₯π₯ + 3)
π₯π₯2β¬
+7π₯π₯β¬
+3π₯π₯β¬
+21
Page 6
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
The question now is: if you knew the answer was π₯π₯2 + 10π₯π₯ + 21, could you figure out that the two binomials that created it were (π₯π₯ + 7)(π₯π₯ + 3)? Letβs see if we can find a pattern that is helpful. Consider these examples:
(π₯π₯ + 7)(π₯π₯ + 3) becomes
π₯π₯2 + 10π₯π₯ + 21
(π₯π₯ + 7)(π₯π₯ β 3) becomes
π₯π₯2 + 4π₯π₯ β 21
(π₯π₯ β 7)(π₯π₯ β 3) becomes
π₯π₯2 β 10π₯π₯ + 21
(π₯π₯ β 7)(π₯π₯ + 3) becomes
π₯π₯2 β 4π₯π₯ + 21 In each result, the first term was π₯π₯2, and the size of the third term was 21. The first column, however, had the 21 as positive while the second column had it as negative. Thatβs because in the first column, both binomial factors had the same sign, while in the second column, one binomial had addition and the other had subtraction. The middle term showed the most variation --- it changed from 10π₯π₯ to 4π₯π₯, and sometimes it was positive while other times it was negative. This goes back to the fact that the middle term is the final result of two multiplications and one addition --- in this case, the addition of 7π₯π₯ and 4π₯π₯, but with varying signs.
1. Copy the following lists of binomial factors and trinomial results. Then match each binomial multiplication with the trinomial that would be its result.
Binomial factors Trinomial results
(π₯π₯ + 5)(π₯π₯ + 2)
(π₯π₯ + 5)(π₯π₯ β 2)
(π₯π₯ β 5)(π₯π₯ β 2)
(π₯π₯ β 5)(π₯π₯ + 2)
π₯π₯2 β 7π₯π₯ + 10
π₯π₯2 + 7π₯π₯ + 10
π₯π₯2 + 3π₯π₯ β 10
π₯π₯2 β 3π₯π₯ β 10
Example 2 Obviously, we donβt always have the option of comparing all 4 versions of how the factors might interact. Usually, we have only the trinomial and must to break it down without choices to draw from. So how do we describe this pattern in a way that we can apply it to any trinomial?
Page 7
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
We begin by looking at the factors of the first term. In this lesson, we will work exclusively with trinomials whose leading coefficient is 1, so the first term will break apart into single letters, like this:
π₯π₯2 + 5π₯π₯ β 14
(π₯π₯ )(π₯π₯ ) We then turn our attention to the last term, the one without the π₯π₯. If you recall from Example 1, the third term stayed 21 (but was sometimes positive and sometimes negative). We need to break the third term into a pair of factors:
π₯π₯2 + 5π₯π₯ β 14
(π₯π₯ 7 )(π₯π₯ 2) Now, keep in mind that 14 can be created by multiplying either 7 & 2 or 1 & 14. So why do we choose 7 & 2? Well, the middle term is created by addingβ¦ Can you see a way to make 5π₯π₯ out of 1 & 14? No. But out of 7 & 2? Sure, if the 7 is positive and the 2 is negative. So we wrap up the problem by placing those signs in the binomials:
π₯π₯2 + 5π₯π₯ β 14
(π₯π₯ + 7 )(π₯π₯ β 2) And the final answer is: (π₯π₯ + 5)(π₯π₯ β 2).
2. Copy the following problems, then use the process outlined above to finish factoring: a. π₯π₯2 + 8π₯π₯ + 7 factors as (π₯π₯ + 1)(π₯π₯ )
b. π₯π₯2 β 3π₯π₯ β 10 factors as (π₯π₯ β 5)( )
c. π₯π₯2 + 2π₯π₯ β 15 factors as (π₯π₯ β 3)( )
d. π₯π₯2 β 6π₯π₯ + 5 factors as (π₯π₯ β 5)( )
Example 3 Some problems are more complicated than others, depending on how many factor pairs the third term has. For example:
24 can factor using 4 & 6, 3 & 8, 2 & 12, or 1 & 24. 60 can factor using 6 & 10, 3 & 20, 2 & 30, 4 & 15, 5 & 12, or 1 & 60
How will you know which one to pick? Well, itβs a guess & check process, but there are some ways to narrow the choices. For instance, if you need to add to an odd number, youβll need an odd and an even. So if you needed multiply to 24 but add to an odd number, you would have to be 8 & 3 or 1 & 24 because those are the
Page 8
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
only odd/even pairs on the list of factors. In contrast, even/even pairs would add to an even number, so if you needed to multiply to 60 but add to an even number, then you would pick 6 & 10 or 2 & 30 because those are the only even/even pairs on the list of factors for 60. How will you know if you have generated all of the factor pairs possible for a certain end term? Thatβs a little more subtle, but involves breaking the factors down into more factors. Take, for example, 72. Most of us, when asked what multiplies to 72, would probably say 8 times 9. 72: 8 x 9 Keep in mind that 8 is the product of 2 & 4, so if we can divide something by 8, we can also divide by 2 and 4: 72: 8 x 9 2 x ______ 4 x ______ But 9 is the product of 3 & 3, so if we can divide by 9, we can also divide by 3: 72: 8 x 9 2 x ______ 4 x ______ 3 x ______ Once we have filled all the blanks (either by using a calculator or doing mental math), we check to make sure all of the factors of all of the factors are represented: 72: 8 x 9 2 x 36 4 x 18 3 x 24 Are all the factors of 18 on the list? Weβve got 2, 9, 18, and 3, but NOT 6. So we write 6 on the list: 72: 8 x 9 2 x 36 4 x 18 3 x 24 6 x ______ and the blank is 12 Now we check again: are all the factors of 12 on the list? Are all the factors of 24 on there? All the factors of 36? Well, then we have the whole list and are free to start looking for the correct pair to use. Letβs say we wanted to factor: π₯π₯2 β 21π₯π₯ β 72. That would need: (π₯π₯ + 3)(π₯π₯ β 24). Letβs say we wanted to factor: π₯π₯2 + 14π₯π₯ β 72. That would need: (π₯π₯ + 18)(π₯π₯ β 4) Letβs say we wanted to factor: π₯π₯2 β 18π₯π₯ + 72. That would need: (π₯π₯ β 6)(π₯π₯ β 12).
Page 9
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
3. Copy each trinomial into your notes, then factor into two binomials. a. π₯π₯2 + 4π₯π₯ β 60
b. π₯π₯2 β 17π₯π₯ + 60
c. π₯π₯2 β 2π₯π₯ β 24
d. π₯π₯2 + 11π₯π₯ + 24 Example 4 What if the middle term is just π₯π₯? Letβs take, for example, π₯π₯2 + π₯π₯ β 72. What that really says is: π₯π₯2 + 1π₯π₯ β 72. We would need a pair of factors that multiplies to 72 but adds to 1. In other words, the numbers must be back-to-back, like 4 & 5 or 6 & 7 or 8 & 9. Did you see one of those on the list from Example 3? Yep. 8 & 9. So to factor π₯π₯2 + π₯π₯ β 72 we would write: (π₯π₯ β 8)(π₯π₯ + 9). Does it matter what order? No. We could write: (π₯π₯ + 9)(π₯π₯ β 8) and it would still be accurate. However, if we write: (π₯π₯ β 9)(π₯π₯ + 8), that would not work because the sign on the 1π₯π₯ would not match the trinomial we were trying to factor. The common mistake in problems like this is to assume that because it has to add to 1, it needs to involve 1. But the reality is that the smaller the middle term becomes, the closer together the two factors need to be. So when you need to add to 1, it can sometimes be easier to list all the pairs of back-to-back numbers (2 & 3, 3 & 4, 4 & 5, etc) until you find the pair that multiplies to the 3rd term of the trinomial.
4. Copy each trinomial into your notes, then factor into two binomials: a. π₯π₯2 β π₯π₯ β 6
b. π₯π₯2 + π₯π₯ β 12
c. π₯π₯2 β π₯π₯ β 20
d. π₯π₯2 + π₯π₯ β 30 What if you have no middle term at all? Take, for example: π₯π₯2 β 25. We saw in the last lesson that this is called a βdifference of two squaresβ and it comes from multiplying conjugate pairs (binomials that use the same numbers, but opposite signs). That means π₯π₯2 β 25 factors as (π₯π₯ + 5)(π₯π₯ β 5). In other words, we have to use a pair of numbers that multiplies to 25 and adds to zero.
5. Copy each trinomial, then factor it: a. π₯π₯2 β 16 b. π₯π₯2 β 49
Page 10
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I
Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Problem Set Factor these trinomials as the product of two binomials.
1. π₯π₯2 β 8π₯π₯ β 20 2. π₯π₯2 β 8π₯π₯ β 33
3. π₯π₯2 β 8π₯π₯ β 9 4. π₯π₯2 β 8π₯π₯ + 12
5. π₯π₯2 + 7π₯π₯ + 12 6. π₯π₯2 + 13π₯π₯ + 12
7. π₯π₯2 + 4π₯π₯ β 12 8. π₯π₯2 + 4π₯π₯ β 5
9. π₯π₯2 + 4π₯π₯ β 32 10. π₯π₯2 + 4π₯π₯ β 45
11. π₯π₯2 + 4π₯π₯ β 60 12. π₯π₯2 + 11π₯π₯ β 60
13. π₯π₯2 β 7π₯π₯ β 60 14. π₯π₯2 β 16π₯π₯ + 60
15. π₯π₯2 β 23π₯π₯ + 60 16. π₯π₯2 + 17π₯π₯ + 60
17. π₯π₯2 + 10π₯π₯ + 25 18. π₯π₯2 β 16π₯π₯ + 64
19. π₯π₯2 + 4π₯π₯ + 4 20. π₯π₯2 β 6π₯π₯ + 9
21. π₯π₯2 + π₯π₯ β 90 22. π₯π₯2 + π₯π₯ β 42
23. π₯π₯2 β π₯π₯ β 56 24. π₯π₯2 β π₯π₯ β 2
25. π₯π₯2 β 81 26. π₯π₯2 β 100
27. π₯π₯2 β 36 28. π₯π₯2 β 1
Page 11
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 3 ALGEBRA I
Lesson 3: Factoring when π β 1 Last edited on 12/30/17
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Lesson 3: Factoring with a Leading Coefficient
Example 1 In Lesson 2, we saw that factoring is the reverse process of multiplication.
Consider the following example of multiplication:
(π₯ + 3)(π₯ + 5) π₯2 + 5π₯ + 3π₯ + 15 π₯2 + 8π₯ + 15.
When we compare the numbers in the factored form with the numbers in the expanded form, we see
that 15 is the product of the two numbers (3 β 5), and 8 is their sum (3 + 5). This is even more obvious
when we look at the expanded form before the like terms are combined.
Now compare the expansion of this binomial product to the one above:
(2π₯ + 3)(4π₯ + 5) 8π₯2 + 10π₯ + 12π₯ + 15 8π₯2 + 22π₯ + 15.
Because of the coefficients on the π₯βs in the binomials, the first term of the trinomial answer is no
longer simply π₯2. However, you can see that the last term, 15, is still the product of 3 β 5.
However, we are no longer βmultiplying to the end and adding to the middleβ. For one thing, there is
no pair of numbers that would multiply to 15 and add to 22.
In the expression between the two arrows, before the like terms were combined, we can see the
coefficients of the linear terms (the ones with the π₯βs). In this case, we have 10π₯ and 12π₯. These
linear terms are not just the sum of the factors of 15 because each factor of 15 interacted with the
factors of the 8π₯2.
The moral of the story is: when there are coefficients in the binomials, the middle terms are no longer
a simple sum.
1. Copy and complete the multiplications shown below in your notes.
a. (5π₯ β 3)(2π₯ + 1)
b. (5π₯ β 3)(2π₯ β 1)
c. (3π₯ β 4)(2π₯ β 7)
d. (3π₯ β 4)(π₯ β 7)
By the end of this lesson, you should be able to:
factor a trinomial whose leading coefficient is a prime number using guess & check
factor a trinomial whose leading coefficient is a prime number using a table
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 3 ALGEBRA I
Lesson 3: Factoring when π β 1 Last edited on 12/30/17
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Example 2 Since the presence of the coefficients in the binomials makes such a difference in the middle term of
the trinomial, how do we then predict what numbers to use when we try to factor a trinomial with a
leading coefficient?
The answer depends a lot on whether you prefer guess & check, which relies fairly heavily on mental
math, or whether you prefer a process with set steps, but the steps have to be memorized.
Letβs take a look at guess & check first. If we wanted to factor this trinomial:
3π₯2 β 13π₯ β 10
We would start by breaking apart the first term:
3π₯2 β 13π₯ β 10
(3π₯ )(π₯ )
You can see thatβs not too different than the factoring from the last lesson. We just need to include the
leading coefficient in the leading binomial. (Well, actually, it could be either binomial.)
Next, we look at the end number: 8 factors as 2 & 4 or 1 & 8. Obviously, none of those pairs adds to 10.
But even if they did, that still wouldnβt necessarily mean that we use them.
Instead, we have to consider their possible placements, and what effect that will have on the middle
term when we multiply the binomials together. Here are the options:
If it isβ¦ The outside pair isβ¦ the inside pair isβ¦ so the middle could beβ¦
(3π₯ 2)(π₯ 5) 15π₯ 2π₯ 13π₯ or β13π₯
17π₯ or β17π₯
(3π₯ 5)(π₯ 2) 6π₯ 5π₯ 11π₯ or β11π₯
1π₯ or β1π₯
Since we want a middle term of β13π₯ for this particular trinomial, then we choose an arrangement
that puts the 3 and 5 as the outside pair, like this:
3π₯2 β 13π₯ β 10
(3π₯ 2 )(π₯ 5 )
The last thing to do is decide on the signs for each binomial. We can only change signs on the 2 & 5,
not on the 3π₯ and π₯. We want a middle term of β13π₯, which will only be possible if the 3π₯ and 5
create β15π₯, and the 2 and π₯ create +2π₯. So we end up with this:
3π₯2 β 13π₯ β 10
(3π₯ + 2 )(π₯ β 5 )
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 3 ALGEBRA I
Lesson 3: Factoring when π β 1 Last edited on 12/30/17
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This method relies heavily on the ability to visualize what the outside/inside combination is going to
be. Itβs a guess & check method, where we guess where the numbers need to go, then check the
outside/inside combination to see if it matches the middle term we need. If it does, then weβre done. If
it doesnβt, we switch the order and try again, or switch to a different pair of numbers.
There are some hints that we can take advantage of. For example, the two numbers inside the
binomials canβt share any common terms. So while we can have (3π₯ + 2), we canβt have (3π₯ + 6).
Also, the signs of the trinomial can help us with the signs of the binomials. If the middle term is
negative and the third term is positive, we know both binomials have subtraction, because we need to
multiply to a positive but add to a negative. If the middle term and third term are both positive, then
we know both binomials have addition.
2. Copy the trinomials into your notes and complete the factorization:
a. 2π₯2 β 5π₯ β 12
(2π₯ )(π₯ 4)
b. 2π₯2 + 9π₯ β 18
(2π₯ 3)(π₯ )
c. 3π₯2 + 10π₯ + 8
(3π₯ 4)(π₯ )
d. 3π₯2 β 25π₯ + 8
(3π₯ )(π₯ 8)
3. Take a look back at letter D. What is it about the middle term that determines the use of 8 & 1
and the placement of the 8?
Example 3 So whatβs the alternative if you donβt want to guess and check?
This method goes back to the idea of the table method for multiplying binomials.
Take, for example, this problem from Lesson 1:
π₯ β5
3π₯ 3π₯2 β15π₯
β2 β2π₯ +10
If we wanted, we could work backwards from the trinomial to the table. It would start like this:
3π₯2 β 17π₯ + 10
3π₯2 β 17π₯ + 10
3π₯2
+10
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 3 ALGEBRA I
Lesson 3: Factoring when π β 1 Last edited on 12/30/17
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The next step is to figure out that the remaining two spots in the table should be.
We know the middle term is supposed to be βπππ, but we need to split that into two separate terms.
How will we know if itβs βπππ and β ππ, or β πππ and β ππ, or β πππ and ππ?
Where this method is trying to go is to be able to figure out what the numbers were at the top of the
columns and the start of the rows. So there needs to be something in common between the πππ and
the two empty space around it, for example. But there also needs to be something in common between
the 10 and the two empty spaces.
To figure out a pair of numbers that would have something in common with both πππ and 10, what
we do is combine them --- specifically, we multiply them --- and then use that product to try and find a
pair of numbers to put into the empty spaces of the table.
So, πππ times 10 is ππππ, and we need to come up with a pair of numbers that multiplies to ππ and
adds to βππ. The pair that makes that work is 15 & 2:
Now that we have the table filled, we look across each row to see what those terms have in common:
Next, we look down each column to see what those terms have in common:
We put the common items in the binomials: (ππ π)(π π)
(Notice that one binomial is the rowsβ common terms, and the other is the columnsβ common terms.)
And the last thing to do is decide on the signs: (ππ β π)(π β π)
We know that they both have to be subtraction because the two terms we picked to fill the table were
βπππ and βππ. That will only happen if both the 2 and the 5 have subtraction in front of them.
3π₯2 β 17π₯ + 10
3π₯2 β15π₯
β2π₯ +10
3π₯2 β 17π₯ + 10
3π₯2 β15π₯
β2π₯ +10
3π₯
2
3π₯2 β 17π₯ + 10
3π₯2 β15π₯
β2π₯ +10
π₯ 5
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 3 ALGEBRA I
Lesson 3: Factoring when π β 1 Last edited on 12/30/17
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Problem Set
Factor the following quadratic expressions.
1. 3π₯2 + 27π₯ + 60 2. 2π₯2 + 9π₯ β 5
3. 3π₯2 β 2π₯ β 5 4. 2π₯2 + 7π₯ β 4
5. 2π₯2 + 5π₯ + 2 6. 5π₯2 + 19π₯ β 4
7. 3π₯2 β 13π₯ + 12 8. 5π₯2 β 7π₯ + 2
9. 4π₯2 + 9π₯ + 2 10. 4π₯2 β π₯ β 3
11. 6π₯2 β 5π₯ + 1 12. 6π₯2 + π₯ β 1
13. 6π₯2 + π₯ β 2 14. 6π₯2 + 11π₯ β 2
15. 8π₯2 β 2π₯ β 3 16. 8π₯2 β 14π₯ + 3
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 4 ALGEBRA I
Lesson 4: Special products & factors Last edited on 12/30/17
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Lesson 4: Special Products and Factors
Example 1 Compare these two problems:
(π₯ β 8)(π₯ + 2)
π₯2 β 8π₯ + 2π₯ β 16
π₯2 β 6π₯ β 16
(π₯ β 4)(π₯ + 4)
π₯2 + 4π₯ β 4π₯ β 16
π₯2 β 16
Notice that both β8 & 2 and β4 & 4 multiply to β16. But these two problems have different results
because of the way the factor pairs combine. In problem A, the β8 & 2 create like terms that are not
the same size, and combine to make β6π₯. But in problem B, because the like terms are equal amounts
but opposite signs, they combine to 0π₯, which drops out of the final answer.
The binomials are known as a conjugate pair and the resulting quadratic is a difference of two squares.
CONJUGATE PAIR: a pair of binomials that have the same terms, but one binomial uses addition and
the other uses subtraction.
Example: (3π₯ β 5)(3π₯ + 5) is a conjugate pair but (3π₯ β 5)(π₯ + 5) is not
(π₯ + 2)(π₯ β 2) is a conjugate pair but (π₯ β 2)(π¦ + 2) is not
DIFFERENCE OF TWO SQUARES: a quadratic binomial (two terms, and the highest power is 2) where
each term is a perfect square and the sign between them is subtraction.
Example: 9π₯2 β 25 is a difference of two squares but 8π₯2 β 25 is not
16π₯2 β 1 is a difference of two squares but 16π₯2 + 1 is not
Here are a few more examples of the same pattern:
(3π₯ β 2)(3π₯ + 2) becomes
9π₯2 + 6π₯ β 6π₯ β 4 and then:
9π₯2 β 4
(π₯ β 7)(π₯ + 7) becomes
π₯2 + 7π₯ β 7π₯ β 49 and then
π₯2 β 49
(2π₯ + 5)(2π₯ β 5) becomes
4π₯2 + 10π₯ β 10π₯ β 25 and then
4π₯2 β 25
In each case, the combination of like terms cancels to zero. So if we see that a problem is multiplying a
conjugate pair, we can skip the Outside and Inside pairs and go directly to the final answer.
1. Copy each multiplication in your notes. Use the shortcut to find the final answer.
a. (π₯ + 9)(π₯ β 9) b. (4π₯ β 7)(4π₯ + 7)
By the end of this lesson, you should be able to:
identify and factor a difference of two squares
identify and factor a perfect square trinomial
A B
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 4 ALGEBRA I
Lesson 4: Special products & factors Last edited on 12/30/17
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Example 2 The flip side of being able to go directly from a conjugate pair to a difference of two squares is that we
can also go directly from a difference of two squares back to the conjugate pair of factors.
For example:
π₯2 β 81 πππππππ β (π₯ + 9)(π₯ β 9)
Of course, with this problem, we could also ask: βwhat multiplies to 81 and adds to zero?β
and that would generate the exact same set of binomials.
Where this pattern is necessary is on problems where we have a leading coefficient that is a perfect
square other than 1:
25π₯2 β 81 πππππππ β (5π₯ + 9)(5π₯ β 9)
And we donβt have to go through any complicated factoring schemes. We just have to break up the 25
and the 81 into pairs of equal numbers, and then have one binomial with addition and one with
subtraction.
2. Copy the following into your notes. Then factor each into conjugate pairs.
a. 49π₯2 β 4 b. 36π₯2 β 1
Example 3 Compare these two problems:
(3π₯ β 4)(π₯ + 4)
becomes:
3π₯2 + 12π₯ β 4π₯ β 16
and then:
3π₯2 + 8π₯ β 16
(3π₯ β 4)(3π₯ β 4)
becomes:
9π₯2 β 12π₯ β 12π₯ + 16
and then:
9π₯2 β 24π₯ + 16
There are several significant differences between these two problems and example 1. First of all,
neither problem C nor problem D are conjugate pairs. Secondly, because of that, neither of them
resulted in a βdifference of two squaresβ kind of answer, even though both have a subtraction and
involve at least one perfect square.
But there are also significant differences between problem C and problem D. In problem C, while the
two binomials both involve a 4, the rest of the binomials are different. On the other hand, in problem
D, it is the same binomial exactly. Weβve seen this before, in lesson 1. Having the same binomial twice
means that you are squaring that binomial: (3π₯ β 4)(3π₯ β 4) πππ ππ π€πππ‘π‘ππ ππ (3π₯ β 4)2.
C D
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 4 ALGEBRA I
Lesson 4: Special products & factors Last edited on 12/30/17
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We discussed in Lesson 1 how to use the resulting pattern to predict the results of multiplying out the
squared binomial. For example:
(4π₯ β 7)2 πππππππ 16π₯2 β 56π₯ + 49
This type of result is often referred to as a perfect square trinomial.
PERFECT SQUARE TRINOMIAL: a three-term quadratic whose first and last terms are perfect squares,
and whose middle term is the product of the square roots of those terms, doubled.
Example: 16π₯2 β 56π₯ + 49 is a perfect square trinomial but 15π₯2 β 56π₯ + 49 is not
π₯2 + 12π₯ + 36 is a perfect square trinomial but π₯2 + 12π₯ β 36 is not
4π₯2 β 12π₯ + 9 is a perfect square trinomial but 4π₯2 β 6π₯ + 9 is not
The question becomes: can we use this binomial squared pattern to predict the results of factoring a
perfect square trinomial?
Here is an example of how that would work:
Given the problem: 9π₯2 β 24π₯ + 16
This problem factors as:
(3π₯ β 4)2
Letβs see another example:
Given the problem: 25π₯2 + 20π₯ + 4
This problem factors as:
(5π₯ + 2)2
3. Copy each quadratic into your notes, then factor it using the perfect square trinomial pattern:
a. 49π₯2 β 14π₯ + 1 b. 4π₯2 + 36π₯ + 81
CAUTION: This works ONLY in the situations described here. If either the first or last terms are not
perfect squares, or if the middle term is not the correct amount, then this style of factoring wonβt work.
4π₯, squared β7, squared 4π₯ times β7 times 2
3π₯ squared β4 squared 3π₯ times β4 times 2
5π₯ squared 2 squared 5π₯ times 2 times 2
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 4 ALGEBRA I
Lesson 4: Special products & factors Last edited on 12/30/17
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Problem Set
Factor each quadratic using the patterns discussed in this lesson:
1. 25π₯2 β 9 2. 100π₯2 β 81
3. 36π₯2 β 1 4. π₯2 β 49
5. 64π₯2 β 25 6. 64π₯2 β 25π¦2
7. 100π₯2 β 60π₯ + 9 8. 16π₯2 + 40π₯ + 25
9. 4π₯2 + 28π₯ + 49 10. 64π₯2 β 16π₯ + 1
11. π₯2 + 10π₯ + 25 12. π₯2 + 10π₯π¦ + 25π¦2
Explain why each quadratic is not factorable with the patterns from this lesson:
13. π₯2 + 5π₯ + 25
14. π₯2 + 10π₯ β 25
15. 2π₯2 β 20π₯ + 25
16. 9π₯2 β 24π₯ + 4
17. 9π₯2 β 24π₯ + 15
18. 16π₯2 + 1
19. 4π₯2 β 3
20. 4π₯2 β π₯ β 9
Page 20
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 ALGEBRA I
Lesson 5: More Factoring Last edited on 7/17/17
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Lesson 5: More Factoring
Example 1 Compare these two problems:
π₯π₯2 β 9π₯π₯ β 10 (π₯π₯ β 10)(π₯π₯ + 1)
2π₯π₯2 β 9π₯π₯ + 10 (2π₯π₯ β 5)(π₯π₯ β 2)
These two problems have different factors because of the way the leading coefficient, 2, interacts with the rest of the factorization. One way to visualize this interaction is by using a process that is sometimes referred to as βdiamond and boxβ. Youβve seen the box before, in the last lesson --- itβs the table that we used to break the middle term into two separate partial sums. The diamond is a way to help visualize the βoutside/insideβ combination of the FOIL process. If you were to use the FOIL process to distribute (2π₯π₯ β 5)(π₯π₯ β 2) you would end up with:
2π₯π₯2 β 4π₯π₯ β 5π₯π₯ + 10 Looking at these four terms, consider this interesting pattern:
2π₯π₯2 π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ + 10 equals 20π₯π₯2 and
β4π₯π₯ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 5π₯π₯ equals 20π₯π₯2 The same pattern is true in example A, also: (π₯π₯ β 10)(π₯π₯ + 1) becomes π₯π₯2 + 1π₯π₯ β 10π₯π₯ β 10 when distributed.
π₯π₯2 π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 10 equals β10π₯π₯2 and
+1π₯π₯ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 10π₯π₯ equals β10π₯π₯2 Here are a few more examples of the same pattern:
(3π₯π₯ β 2)(2π₯π₯ + 7) becomes
6π₯π₯2 + 21π₯π₯ β 4π₯π₯ β 14 and
6π₯π₯2 π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 14 π‘π‘π‘π‘ β 84π₯π₯2 +21π₯π₯ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 4π₯π₯ π‘π‘π‘π‘ β 84π₯π₯2
(3π₯π₯ β 2)(π₯π₯ + 7) becomes
3π₯π₯2 + 21π₯π₯ β 2π₯π₯ β 14 and
3π₯π₯2 π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 14 π‘π‘π‘π‘ β 42π₯π₯2 +21π₯π₯ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 2π₯π₯ π‘π‘π‘π‘ β 42π₯π₯2
(π₯π₯ β 2)(π₯π₯ + 7) becomes
π₯π₯2 + 7π₯π₯ β 2π₯π₯ β 14 and π₯π₯2 π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 14 π‘π‘π‘π‘ β 14π₯π₯2
+7π₯π₯ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ β 2π₯π₯ π‘π‘π‘π‘ β 14π₯π₯2
By the end of this lesson, you should be able to: β’ factor a trinomial whose leading coefficient is not prime β’ identify and factor out a greatest common factor β’ factor a trinomial into a greatest common factor monomial and two binomials
A B
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 ALGEBRA I
Lesson 5: More Factoring Last edited on 7/17/17
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As you can see, in each case, the product of the First and Last terms equals the product of the Outside and Inside terms. We can then use this pattern to predict the Outside/Inside terms, because we know they will have to be factors of the product of the First/Last terms. For example:
Given this: then the First/Last product is:
So the Outside/Inside terms need to:
and that means they must be:
6π₯π₯2 β 17π₯π₯ + 5 30π₯π₯2 multiply to 30π₯π₯2 and add to β17π₯π₯ β15π₯π₯ and β2π₯π₯
10π₯π₯2 β π₯π₯ β 3 β30π₯π₯2 multiply to β30π₯π₯2 and add to β1π₯π₯ 5π₯π₯ and β6π₯π₯
15π₯π₯2 + 4π₯π₯ β 4 β60π₯π₯2 multiply to β60π₯π₯2 and add to +4π₯π₯ 10π₯π₯ and β6π₯π₯
So, where does a diamond play into all of this? Well, we sometimes use that structure to help organize our thoughts. Hereβs how it would look for the problem: 15π₯π₯2 + 4π₯π₯ β 4 15π₯π₯2 + 4π₯π₯ β 4 Once the diamond is filled, we move on to the box: 15π₯π₯2 + 4π₯π₯ β 4 And from the box, we move on to the factors:
β60π₯π₯2
10π₯π₯ β6π₯π₯
+4π₯π₯
β’ Multiply first and last to fill the top corner β’ Write middle term in the bottom corner β’ Fill the two βsideβ corners with the #s that
multiply to top and add to bottom
β60π₯π₯2
10π₯π₯ β6π₯π₯
+4π₯π₯
15π₯π₯2 β6π₯π₯
10π₯π₯ β4
15π₯π₯2 β6π₯π₯
10π₯π₯ β4
Shared: 3π₯π₯
2
Shared: 5π₯π₯ β2
Final result:
15π₯π₯2 + 4π₯π₯ β 4
Factors as:
(3π₯π₯ + 2)(5π₯π₯ β 2)
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 ALGEBRA I
Lesson 5: More Factoring Last edited on 7/17/17
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1. Copy each trinomial into your notes, then use the diamond & box structure to factor each one: a. 10π₯π₯2 β π₯π₯ β 3 b. 6π₯π₯2 β 17π₯π₯ + 3
Example 2 We started our factoring discussion by looking at problems with a leading coefficient of 1 (for example, π₯π₯2 β 3π₯π₯ β 10). We then expanded the discussion by investing problems with a leading coefficient that is prime (for example, 2π₯π₯2 β 9π₯π₯ + 10). And this lesson covered one way to factor when the leading coefficient is not prime, or in other words, composite (for example, 6π₯π₯2 β 17π₯π₯ + 3). However, there are trinomials with a leading coefficient other than 1 where it is NOT to your advantage to use a diamond & box approach. These problems are ones where the leading coefficient is a shared factor between all of the terms of the trinomial. Compare these two problems:
2π₯π₯2 β 9π₯π₯ + 10 2π₯π₯2 β 8π₯π₯ β 10
Both of these examples have a leading coefficient that happens to be prime. But problem Dβs leading coefficient is also a factor of the other two terms in the problem. This concept of a shared factor goes back to elementary school, when you first learned about the Greatest Common Factor between two numbers (for example, the GCF of 12 and 18 is 6). Here, the 2 that is the leading coefficient is also the GCF between 2, 8, and 10. Removing the GCF looks like this: 2π₯π₯2 β 8π₯π₯ β 10 πππ‘π‘πππππ‘π‘π‘π‘π‘π‘ 2(π₯π₯2 β 4π₯π₯ β 5) You are factoring, which means you are breaking up multiplication, and in this case, that means dividing both the 8 and the 10 by 2. Once the GCF is removed, it turns out that the remaining trinomial, π₯π₯2 β 4π₯π₯ β 5, can factor further. There are numbers that multiply to 5 and add to 4. So the entire process looks like this:
2π₯π₯2 β 8π₯π₯ β 10 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½2(π₯π₯2 β 4π₯π₯ β 5)
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½2(π₯π₯ β 5)(π₯π₯ + 1)
2. Copy each trinomial into your notes, then factor out the GCF of each.
a. 3π₯π₯2 + 5π₯π₯ β 42 b. 4π₯π₯2 β 28π₯π₯ + 24
3. After factoring out the GCF, fully factor the remaining trinomial into two binomials. CAUTION: You could factor these problems without removing the GCF first. For example, 2π₯π₯2 β 8π₯π₯ β10 could become (2π₯π₯ β 10)(π₯π₯ + 1). However, you would then have to remove the GCF from the first parentheses in order to meet the βfully factoredβ requirement: 2(π₯π₯ β 5)(π₯π₯ + 1).
C D
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 ALGEBRA I
Lesson 5: More Factoring Last edited on 7/17/17
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Example 3 Here is an example of each type of factoring discussed so far in Lessons 1 - 4:
π₯π₯2 + 8π₯π₯ β 20 π₯π₯2 + π₯π₯ β 20 π₯π₯2 β 25
2π₯π₯2 β 3π₯π₯ β 5 6π₯π₯2 + 7π₯π₯ β 5 3π₯π₯2 β 18π₯π₯ + 15 Each of these problems shares two common characteristics --- they all begin with a quadratic term (power of 2) and end with a constant (a number without a variable). They all have different middle terms (consider π₯π₯2 β 25 as having a middle term of 0π₯π₯) and different leading coefficients, and each takes a slightly different approach when it comes to breaking the trinomial back down into the binomials used to create it. However, there is one version of a quadratic that doesnβt have these two common characteristics. It doesnβt come from multiplying from two binomials, but rather from multiplying a monomial and a binomial. Compare these two problems:
(2π₯π₯ β 5)(π₯π₯ β 2) becomes
2π₯π₯2 β 4π₯π₯ β 5π₯π₯ + 10 which becomes 2π₯π₯2 β 9π₯π₯ + 10
2π₯π₯(π₯π₯ β 5) becomes
2π₯π₯2 β 10π₯π₯ which doesnβt have a trinomial form
Notice that problem F is not the same style as the two-term quadratic π₯π₯2 β 25, which is missing a middle term. Instead, F is missing its constant --- there is no βnumber without an π₯π₯β to end the problem. And since factoring often involves the discussion of βwhat multiplies to the end and adds to the middleβ, not having a constant term is going to create issues. So how do we factor a quadratic like this? Well, we have to consider not βwhat multiplies to the end, etcβ, but rather, what do the terms share --- what is their Greatest Common Factor? Take, for example, the two-term quadratic: 15π₯π₯2 + 10π₯π₯. Letβs break each term into its parts:
15π₯π₯2 π‘π‘π‘π‘ 3 β 5 β π₯π₯ β π₯π₯ 10π₯π₯ π‘π‘π‘π‘ 2 β 5 β π₯π₯
Whatβs circled are the shared items, and putting those items together creates the GCF: 5π₯π₯ Like we did in example 2, we then factor out the GCF. What isnβt circled gets left behind as a single binomial:
15π₯π₯2 + 10π₯π₯ ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½5π₯π₯(3π₯π₯ + 2)
E F
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 5 ALGEBRA I
Lesson 5: More Factoring Last edited on 7/17/17
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4. Copy each quadratic into your notes, then factor out the GCF of each to leave an answer that is fully factored:
a. 8π₯π₯2 β 12π₯π₯ b. 10π₯π₯2 + 20π₯π₯ c. β6π₯π₯2 β 12π₯π₯ (hint: factor out β6π₯π₯) d. βπ₯π₯2 + 3π₯π₯ (hint: factor out βπ₯π₯)
Problem Set Fully factor each quadratic:
1. 5π₯π₯2 β 30π₯π₯ 2. 20π₯π₯2 + 12π₯π₯
3. 25π₯π₯2 β 30π₯π₯ 4. 24π₯π₯2 + 12π₯π₯
5. π₯π₯2 β 10π₯π₯ 6. π₯π₯2 β π₯π₯
7. 10π₯π₯2 β 20π₯π₯ β 30 8. 6π₯π₯2 + 18π₯π₯ + 12
9. 4π₯π₯2 + 8π₯π₯ β 32 10. 3π₯π₯2 β 3π₯π₯ β 18
11. 5π₯π₯2 β 45 12. 3π₯π₯2 β 60
13. 6π₯π₯2 + 5π₯π₯ β 6 14. 5π₯π₯2 β 6π₯π₯ β 8
15. 4π₯π₯2 + 11π₯π₯ + 6 16. 6π₯π₯2 β 11π₯π₯ + 5
Page 25
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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Lesson 6: The Zero Product Property
Example 1 Consider the function: π(π) = ππ + ππ β ππ
Letβs say you wanted to find the value of π(π). That process would look like this:
π(π) = (π)π + π(π) β ππ
π(π) = π + π β ππ
π(π) = βπ
This particular trinomial factors like this: ππ + ππ β ππ πππππππ β (π + π)(π β π)
What if we use the factored version to find the value at 1? It should come out the same as before:
π(π) = (π + π)(π β π)
π(π) = ( π )( βπ )
π(π) = βπ
And it does!
Letβs try this with a few more values:
π(π) = ππ + ππ β ππ π(π) = (π + π)(π β π)
π(π) = (π)π + π(π) β ππ
π(π) = ππ + ππ β ππ
π(π) = ππ
π(π) = (π + π)(π β π)
π(π) = ( ππ )( π )
π(π) = ππ
π(βπ) = (βπ)π + π(βπ) β ππ
π(βπ) = π β ππ β ππ
π(βπ) = βππ
π(βπ) = (βπ + π)(βπ β π)
π(βπ) = ( π )( βπ )
π(βπ) = βππ
1. Copy and complete the following in your notes:
π(π) = ππ + ππ β ππ π(π) = (π + π)(π β π)
π(π) = (π)π + π(π) β ππ
π(π) =
π(π) = (π + π)(π β π)
π(π) =
Of course, the value of π(π) is the y-intercept, so that is a special point.
But there are two other special values that we should examine more closely: π(βπ) and π(π).
By the end of this lesson, you should be able to:
find the π-intercepts of a quadratic function that is factorable
solve a quadratic equation involving a trinomial that is factorable
Page 26
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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π(π) = ππ + ππ β ππ π(π) = (π + π)(π β π)
π(βπ) = (βπ)π + π(βπ) β ππ
π(βπ) = ππ β ππ β ππ
π(βπ) = π
π(βπ) = (βπ + π)(βπ β π)
π(βπ) = ( π )( βπ )
π(βπ) = π
π(π) = (π)π + π(π) β ππ
π(π) = π + π β ππ
π(π) = π
π(π) = (π + π)(π β π)
π(π) = ( π )( π )
π(π) = π
We just found the two points, (βπ, π) and (π, π), that are the π-intercepts of the function. On the
graph, this is where the function will pass through the π-axis. They are also the solutions to the
equation ππ + ππ β ππ = π.
So how can we find these two points without having to guess & check? Take another look at the way
this trinomial factored: ππ + ππ β ππ ππππππ β (π + π)(π β π)
In the factor (π + π), when we plugged in βπ, the factorβs value equaled zero.
In the factor (π β π), when we plugged in π, the factorβs value equaled zero.
In other words, the values of π = βπ and π = π were the solutions to the equations:
π + π = π and π β π = π
This means that to find the π-intercepts of a trinomial, we should factor it, and then look for the values
that would make each factor equal to zero.
Here are a few more trinomials that have been factored, and the solutions to make each factor equal
zero, and the results of substituting those values into the function:
trinomial & factors: solutions: results of substituting:
ππ + ππ β ππ
(π + π)(π β π)
π = βπ and
π = π
(βπ)π + π(βπ) β ππβ ππ β ππ β ππβ π
(π)π + π(π) β ππβ π + π β ππβ π
ππ β ππ + π
(π β π)(π β π)
π = π and
π = π
(π)π β π(π) + πβ π β ππ + π β π
(π)π β π(π) + πβ π β ππ + π β π
ππ β ππ β π
(π β π)(π + π)
π = π and
π = βπ
(π)π β π(π) β πβ ππ β ππ β πβ π
(βπ)π β π(βπ) β π β π + π β πβ π
2. Copy each trinomial into your notes. Factor into two binomials. Then list the values that
would make each binomial equal to zero.
a. ππ β ππ β ππ b. ππ + πππ + ππ
Page 27
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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Example 2 When you were in elementary school, you probably learned that any number, multiplied by zero,
equals zero. Here, we are using the flip side of the same coin --- the knowledge that if a product equals
zero, then one of the two factors involved must be zero. This is called the Zero Product Property.
For a very simple example, consider the equation: ππ = π
If we start with 5, multiply by an unknown number, and get an answer of zero, then the value of the
unknown variable has to be zero. In other words, the solution to the equation ππ = π is that π = π.
If we expand the idea to cover two unknown numbers, it looks like this: ππ = π
And what this would mean is that one or the other of the two variables equals zero: π = π ππ π = π
If we then express these two unknown factors as binomials, it might look like this:
(π + π)(π β π) = π
The fact that the product of the two binomials equals zero means that the value of one or the other of
the binomials equals zero. In other words,
π + π = π ππ π β π = π
Solving these two single-step equations would then lead to the answers:
π = βπ ππ π = π
The idea of solving for what makes each factor equal zero is the same here as it was in Example 1. Itβs
just that the problem is being presented as an equation that needs to be solved, instead of a function
where we need to find the π-intercepts. However, the results of those two processes are the same.
Compare these two problems:
find the π-intercepts for:
π(π) = ππ β ππ β π
(π β π)(π + π)
the first binomial = 0 when π = π
the second binomial = 0 when π = βπ
so the π-intercepts are:
(π, π) and (βπ, π)
solve the equation:
ππ β ππ β π = π
(π β π)(π + π) = π
by the zero product property, this means
π β π = π ππ π + π = π +π + π βπ β π
π = π ππ π = βπ
You can see that in both cases, the values that we find for π are the same.
Sometimes, the equation may be presented out of order, or with extra terms, like this:
ππ β ππ = π
or ππ β ππ β π = π
or πππ β ππ β ππ = ππ β ππ β ππ
In those situations, our first job will be to put the terms on the same side, in the correct order, before
we attempt to solve it.
Page 28
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Remember, the Zero Product Property states that if two items multiply to a value of zero, then one of
the two items equals zero. For our purposes, this means that one side of the equation has to equal
zero, or we canβt proceed with the solution process.
So here are the steps we will take:
Put all terms on one side (keep in mind that we want the ππ term to end up positive)
Combine any like terms
Factor completely
Write each factor with = π after it
Solve for the variable
Here is an example of how this process might look:
Solve the equation:
πππ β ππ + π = πππ + ππ β π βπππ β ππ + π β πππ β ππ + π
ππ β πππ + π = π
(π β π)(π β π) = π
π β π = π ππ π β π = π
+π + π +π + π
π = π or π = π
3. Copy each trinomial into your notes. Put all of the terms on the left side. Factor the trinomial.
Set each factor equal to zero, then solve.
a. ππ + ππ = ππ b. ππ + ππ β π = ππ
Answers to problem set on next page:
1. π = π ππ π = βπ 2. π = π ππ π = π 3. π = ππ ππ π = βπ 4. π = βπ ππ π = π
5. π = π ππ π = βπ 6. π = π ππ π = βπ 7. π = π ππ π = βπ 8. π = π ππ π = βπ
9. π = βπ ππ π = π 10. π = π ππ π = π 11. π = βππ ππ π = βπ 12. π = βππ ππ π = βπ
13. π = βπ ππ π = π 14. π = βπ ππ π = π 15. π = π ππ π = π 16. π = βπ ππ π = βπ
Page 29
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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Problem Set
Factor and solve the following equations. Show all work.
1. π₯2 β 3π₯ β 18 = 0 2. π₯2 β 9π₯ + 8 = 0
3. π₯2 β 4π₯ β 40 = 0 4. π₯2 + π₯ β 30 = 0
5. π₯2 + 5π₯ β 14 = 0 6. π2 + 4π β 5 = 0
7. π2 β 10π + 16 = 0 8. π2 β 7π β 8 = 0
9. π₯2 β 10π₯ + 25 = 0 10. π₯2 + 8π₯ + 16 = 0
Page 30
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I
Lesson 6: Zero Product Property Last edited on 12/30/17
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Put all the terms on the left side, then factor and solve. Show all work.
11. π2 β 7π β 8 = 10 12. π2 β 7π β 8 = β5π
13. π2 β 7π β 8 = β2π + 16 14. π₯2 + 2π₯ = 3
15. π₯2 + 2 = 3π₯ 16. π2 β 9π + 10 = 2π
17. π2 + 15π + 40 = 4 18. π2 + 5π β 35 = 3π
19. 6π2 β 11π = 5π2 β 18 20. 2π₯2 + 11π₯ = π₯2 β π₯ β 32
Page 31
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 7 ALGEBRA I
Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 7: The Zero Product Property, Day 2
Example 1 Consider the function: ππ(ππ) = ππππ + ππππ β ππππ This particular trinomial factors like this: ππππ + ππππ β ππππ
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ (ππ + ππ)(ππ β ππ)
In the last lesson, we discussed how the points, (βππ,ππ) and (ππ,ππ) are the ππ-intercepts of the function, and that they are also the solutions to the equation ππππ + ππππ β ππππ = ππ. So ππ-intercepts are the same as the ππ values at which the function equals zero. We solve that equation using this process:
β’ Put all terms on one side (keep in mind that we want the ππππ term to end up positive) β’ Combine any like terms β’ Factor completely β’ Write each factor with = ππ after it β’ Solve for the variable
This particular trinomial has a leading coefficient of 1, and factors into two binomials. But weβve had several lessons with trinomials that required different types of factoring. So the question becomes: How does this solution process transfer to these other types of trinomials? Compare these two problems:
ππππ β ππππ β ππππ = ππ factors as:
(ππ β ππ)(ππ + ππ) = ππ solves as:
ππ = ππ ππππ ππ = βππ
ππππ β ππππ = ππ factors as:
(ππ β ππ)(ππ + ππ) = ππ solves as:
ππ = ππ ππππ ππ = βππ
As you can see, the only thing that changes here is the fact that problem B doesnβt have a middle term. That affects the choice of factors, but nothing else.
1. Copy each problem into your notes. Then solve each problem according to the steps in the bulleted list shown above:
a. ππππ β ππππ = ππ b. ππππ β ππππ = ππππ
By the end of this lesson, you should be able to: β’ find the ππ-intercepts of a quadratic function that is factorable β’ solve a quadratic equation involving a trinomial that is factorable
A B
Page 32
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 7 ALGEBRA I
Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 2 There are two types of problems where factoring out a Greatest Common Factor becomes necessary. Here is an example of each type:
ππππππ + ππππππ β ππππ = ππ factors as:
πποΏ½ππππ + ππππ β πποΏ½ = ππ and then:
ππ(ππ + ππ)(ππ β ππ) = ππ
ππππππ + ππππππ = ππ factors as:
ππππ(ππ+ ππ) = ππ
You can see that in problem C, the greatest common factor was a constant, 2, while in problem D, the constant was a monomial, ππππ. That is going to affect the way that the two problems continue. Compare the next few steps:
ππππππ + ππππππ β ππππ = ππ πποΏ½ππππ + ππππ β πποΏ½ = ππ ππ(ππ + ππ)(ππ β ππ) = ππ
set each factor equal to zero: ππ = ππ ππ + ππ = ππ ππ β ππ = ππ
ππππ ππππππππππππππππ βππ β ππ +ππ + ππ ππ = βππ ππ = ππ
ππππππ + ππππππ = ππ ππππ(ππ+ ππ) = ππ
set each factor equal to zero:
ππππ = ππ ππ + ππ = ππ Γ· ππ Γ· ππ βππ β ππ ππ = ππ ππ = βππ
Notice that in problem C, where we factored out a constant GCF, no solution came from that constant. But in problem D, the GCF of a coefficient and variable generated a solution. Letβs see that again, with some different numbers:
ππππππ + ππππππ+ ππππ = ππ factors as:
πποΏ½ππππ + ππππ + πππποΏ½ = ππ and then:
ππ(ππ + ππ)(ππ + ππ) = ππ so the next step is:
ππ = ππ ππππ ππ + ππ = ππ ππππ ππ + ππ = ππ which solves as:
ππ = βππ ππππ ππ = βππ (since ππ = ππ doesnβt have a solution,
we can ignore that part)
βππππππ + ππππππ = ππ factors as:
βππππ(ππ β ππ) = ππ so the next step is:
βππππ = ππ ππππ ππ β ππ = ππ which solves as:
ππ = ππ ππππ ππ = ππ
So the bottom line is this: a GCF thatβs only a constant will not generate any solutions, but a GCF that
C D
E F
C D
Page 33
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 7 ALGEBRA I
Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
includes a variable will contribute an automatic answer of ππ = ππ, regardless of its coefficient.
2. Copy each trinomial into your notes. Put all of the terms on the left side. Factor the trinomial. Set each factor equal to zero, then solve.
a. ππππππ = ππππ + ππ b. ππππ + ππππ β ππ = βππ Example 3 In previous lessons, we have dealt with problems whose leading coefficient is not a GCF, but rather is incorporated within one of the two binomial factors. For example, in the trinomial ππππππ + ππππ β ππ, the leading coefficient is not a factor of each term. So we are unable to remove it as a GCF. Instead, we break up the ππππππ between the two binomial factors, like this:
(ππππ )(ππ ) Then we start to guess & check with the factor pairs of 5, or we use a diamond & box process, to fill in the rest of the binomials. Eventually, we end up with this:
(ππππ + ππ)(ππ β ππ) So, how does that coefficient alter the solution process? The entire thing would look like this:
ππππππ + ππππ β ππ = ππ (ππππ + ππ)(ππ β ππ) = ππ
ππππ + ππ = ππ ππππ ππ β ππ = ππ βππ β ππ +ππ + ππ ππππ = βππ ππππ ππ = ππ
Γ· ππ Γ· ππ ππ = β ππ
ππ
The bottom line here is that when there is a coefficient inside the binomial, that binomialβs solution will be a fraction. (This will always be true, assuming that any GCF is already removed.)
3. Copy each trinomial into your notes. Put all the terms on one side. Factor and solve. a. ππππππ + ππππ = ππππππ b. ππππππ + ππππ + ππ = ππππ β ππππ + ππ
Example 4 Sometimes, the leading coefficient is composite, rather than prime (for example, 6 instead of 3). In that case, you might find yourself with coefficients in both binomials, meaning both answers are fractions. Itβs also possible that you might remove a GCF and still have a leading coefficient other than one.
Page 34
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 7 ALGEBRA I
Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Consider these examples:
ππππππ β ππππ + ππ = ππ factors as:
(ππππ β ππ)(ππππ β ππ) = ππ and solves as:
ππ = ππ ππ
ππππ ππ = ππ ππ
ππππππππ + ππππππ β ππ = ππ factors as:
(ππππ + ππ)(ππππ β ππ) = ππ and solves as:
ππ = βππππ
ππππ ππ = ππ ππ
ππππππ β ππππ β ππ = ππ factors as:
(ππππ + ππ)(ππ β ππ) = ππ and solves as:
ππ = β ππ ππ
ππππ ππ = ππ
ππππππ + ππππππ + ππ = ππ factors as:
πποΏ½ππππππ + ππππ + πποΏ½ = ππ ππ(ππππ + ππ)(ππ + ππ) = ππ
and solves as: ππ = β ππ
ππ ππππ ππ = β ππ
ππ
Notice that the GCF of 2 in the lower left does not contribute a solution to the equation. On top of all of these possibilities, there is also the possibility that the problem may be presented with the terms out of order, or with like terms on each side of the equals sign, or some other condition that requires us to rearrange the equation before beginning the solution process. Remember, what we need in order to factor and solve is:
β’ All the terms on one side, with the ππππ term positive β’ All like terms combined β’ Powers in order from highest to lowest
For example:
ππππππππ + ππππππ = ππ + ππ β ππππππππ +ππππππππ β ππ β ππ β ππ β ππ + ππππππππ
ππππππππ + ππππ β ππ = ππ Now we factor completely: πποΏ½ππππππππ + ππππ β πποΏ½ = ππ
ππ(ππππ + ππ)(ππππ β ππ) = ππ Next, set each factor that involves a variable equal to zero, and solve:
ππππ + ππ = ππ ππππ ππππ β ππ = ππ ππ = βππ
ππ ππππ ππ = ππ
ππ
4. Copy each trinomial into your notes. Solve like the example above.
a. ππππππππ β ππππππ + ππππ = ππ β ππππ b. ππππππππ β ππππ = ππππ + ππ
Page 35
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 7 ALGEBRA I
Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17
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Problem Set Solve the following equations. Show all work. 1. π₯π₯2 β 64 = 0 2. π₯π₯2 β 45 = 4 3. 4π₯π₯2 β 4π₯π₯ β 24 = 0 4. 2π₯π₯2 + 6π₯π₯ β 18 = 2 5. ππ2 β 7ππ = 0 6. 5ππ2 β 13ππ = 2ππ 7. 3ππ2 β 75 = 0 8. 2ππ2 β 32 = 0 9. 4π₯π₯2 + π₯π₯ = 3 10. 5π₯π₯2 + 2 = 3π₯π₯ 11. 3ππ2 + 9ππ + 2 = 2ππ 12. 4ππ2 β 11ππ + 4 = 1 β 3ππ 13. 12π₯π₯2 β 24 = 2π₯π₯ + 6 14. 17ππ2 + 4ππ + 10 = 13 β 5ππ β 13ππ2 15. 6ππ2 β 11ππ = 0 16. 3π₯π₯2 + 2π₯π₯ = 0 Partial answers: 1) 8 2) 7 3) 3 4) 2 5) 0 6) 3 7) 5 8) 4 9) -1 10) 1 11) β2 12) 1
2 13) 5
3 14) 1
5 15) 0 16) β2
3
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 8 ALGEBRA I
Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 8: Solving by Taking the Square Root
Example 1
Because this module focuses mostly on variables that are squared, itβs best to have a thorough
understanding of what it means to square something.
On a physical level, squaring means itβs possible to create a square array out of the amount. For example:
42 = 16 means
16 dots in
4 rows and
4 columns
In comparison, a number like 15 can only be arranged in a rectangular array:
15 dots in
3 rows and or 5 rows and
5 columns 3 columns
Or, possibly, in a partially complete square array:
15 dots in
a 4 by 4 array,
missing one dot
On a mathematical level, squaring something means that you are multiplying it by itself:
42 πππππ 4 Γ 4
This is why 42 and (β4)2 come out to the same result. Both calculations involve exact duplicates:
42 πππππππ β 4 Γ 4 and (β4)2
πππππππ β β4 Γ β4
On the other hand, β42 represents β1 Γ 4 Γ 4.
By the end of this lesson, you should be able to:
solve a quadratic equation by taking the square root
identify when a quadratic can be solved by taking the square root and when it canβt
simplify the square root of a perfect square
estimate the value of a square root of a non-perfect square as between two integers
Page 37
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 8 ALGEBRA I
Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17
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So if that is what squaring means, then what does it mean to βsquare rootβ something?
The following pattern shows the mathematical meaning:
ifβ¦ thenβ¦
(5)2 and (β5)2 both equal 25 the square root of 25 is either 5 or β5
(3)2 and (β3)2 both equal 9 the square root of 9 is either 3 or β3
(7)2 and (β7)2 both equal 49 the square root of 49 is either 7 or β7
(4)2 and (β4)2 both equal 16 the square root of 16 is either 4 or β4
(9)2 and (β9)2 both equal 81 the square root of 81 is either 9 or β9
Squaring and taking the square root are inverse operations --- one undoes the other. But there is an
issue with the potential sign of the square root result.
When mathematicians want the negative answer to a square root, we write this:
ββ25 πππππππ β β 5
If we want the positive version (sometimes referred to as the principal root), we write it like this:
β9 πππππππ β 3
And for those situations where we want both, we write it like this:
Β±β16 πππππππ β Β± 4
This is read out loud as βplus or minus the square root of 16 becomes plus or minus 4β.
Generally speaking, if the square root symbol is already there, then the sign is already chosen. But if
we introduce the square root symbol during a solution process, then we write the plus-or-minus
symbol as well.
The square root symbol is called a radical sign and the number underneath is called the radicand.
Certain radicands are known as perfect squares because their square roots are integers. The first
twelve perfect squares are:
1 4 9 16 25 36 49 64 81 100 121 144
If you were to take the square root of any of these, you would get an integer. For example:
β100 β 10 β β4 β β2 Β± β49 β Β±7
But if you were to take the square root of any numbers in between the perfect squares, you would get
an irrational number --- something whose decimal equivalent is non-terminating (doesnβt stop) and
non-repeating (doesnβt have a pattern). For example:
β99 = 9.949874371066β¦.
Page 38
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 8 ALGEBRA I
Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17
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We can estimate the approximate decimal value of an irrational square root by comparing the two
perfect squares on either side. For example:
β28 is between 5 and 6 because 28 is between 25 and 36
β β8 is between β2 and β3 because 8 is between 4 and 9
1. Copy each square root into your notes. Find the integer value, if possible. If not, estimate the
approximate decimal value.
a. β36
b. ββ144
c. β10
d. β β60
2. Write the list of the first 20 perfect squares in your notes. Use your calculator to find the
values beyond 122.
Example 2
In the previous lesson, we looked at how to solve equations that were factorable.
Consider this problem, for example: π₯2 β 16 = 0
We solved the equation by factoring: (π₯ β 4)(π₯ + 4) = 0
And then solving each factor: π₯ = 4 ππ π₯ = β4
Now, consider the quadratic equation: π₯2 = 16
See how thatβs basically the same equation, but with the 16 on the other side of the equals sign?
We just discussed out that the square root of 16 is either 4 or β4.
So could we solve the equation π₯2 = 16 by taking the square root of 16?
Well, sort of⦠You see, we have to take the square root of both sides, like this:
π₯2 = 16
Β±βπ₯2 = Β±β16
π₯ = Β±4
Remember, when the radical is already there, the sign is decided, but when we write in the radical, we
also have to write in the plus-or-minus sign. Youβll notice that we donβt bother to keep it on the π₯ side.
Itβs already a variable, meaning that it can vary. The plus-or-minus is just part of that variability.
The critical part of this method is that each side must be treated as a single item. Square roots canβt
βdistributeβ. For example:
βπ₯2 β 16 = β0 does NOT become βπ₯2 β β16 = 0 and then π₯ β 4 = 0
Page 39
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 8 ALGEBRA I
Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17
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Here are some examples of equations solved using the square root process, and some equations that
canβt be solved this way. Pay close attention to the steps that can be done and the steps that canβt:
YES NO
π₯2 = 25
Β±βπ₯2 = Β±β25
π₯ = Β±5
π₯2 + 25 = 0
Β±βπ₯2 + 25 = Β±β0
canβt finish --- square roots canβt be
βdistributedβ to multiple terms
π₯2 β 4 = 0 +π + π
π₯2 = 4
Β±βπ₯2 = Β±β4
π₯ = Β±2
π₯2 + 25 = 0 βππ β ππ
π₯2 = β25
Β±βπ₯2 = Β±ββ25
canβt finish --- square roots of negative
radicands canβt be done w/ real numbers
3π₯2 = 12 Γ· π Γ· π
π₯2 = 4
π₯ = Β±2
3π₯2 = 12π₯ Γ· π Γ· π
π₯2 = 4π₯
Β±βπ₯2 = Β±β4π₯
π₯ = Β±2βπ₯
canβt finish with this method ---
not allowed to divide by a variable
Notice in the last example in the βNOβ column that having a variable on each side of the equation
makes the equation unable to be solved with this method. We could solve this equation using the
methods shown in the previous lesson, however. When the equation has both an π₯2 and an π₯, we
canβt square root to get the answer --- we have to factor.
3. Copy and solve each equation in your notes. Show all steps.
a. ππ β ππ = π b. πππ = ππ
Example 3
As you saw in example 2, we can use the properties of equality (adding/subtracting/dividing, even
multiplying if needed, to both sides) to arrange the equation before the square root is applied.
What we need for this method to work is a single squared item on both sides of the equation. But the
item doesnβt have to be just an ππ. For example:
(π + π)π = ππ can become Β± β(π + π)π = Β±βππ
The parentheses around the π + π make it a single item, which means the square root can be applied.
Page 40
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 8 ALGEBRA I
Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17
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Hereβs how the solution process would start:
(π + π)π = ππ
Β±β(π + π)π = Β±βππ
π + π = Β±π βπ β π
π = βπ Β± π
Normally, by the time we have π by itself, the problem is done. Here, however, we have the statement
β3 plus or minus 4β. That needs to be finished --- what is 3 plus 4? what is 3 minus 4? --- and those
results are the final answers.
So the last few steps would look like this:
π = βπ Β± π
π = βπ + π ππ π = βπ β π
π = π ππ π = βπ
4. Copy and solve in your notes:
a. (π₯ + 5)2 = 9 b. (π₯ β 7)2 = 25
Problem Set
Solve each equation.
1. π₯2 = 81
2. π₯2 = 144
3. π₯2 β 9 = 0
4. π₯2 β 1 = 0
5. 2π₯2 = 98
6. 3π₯2 = 48
7. (π₯ + 2)2 = 36
8. (π₯ β 4)2 = 121
9. (π₯ + 3)2 = 100
10. (π₯ β 5)2 = 4
11. Explain why each equation cannot be solved using the square root method from this lesson:
a. π₯2 + 4 = 0 b. π₯2 + 4π₯ = 0
Page 41
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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Lesson9:SimplifyingSquareRoots
Example1 Inthelastlesson,wediscussedwhatitmeanstosquareaquantity,withbothaphysicalrepresentationandwithamathematicalrepresentation.Onaphysicallevel,squaringmeansitβspossibletocreateasquarearrayoutoftheamount.Forexample:
4 16means16dotsin4rowsand4columns
Onamathematicallevel,squaringsomethingmeansthatyouaremultiplyingitbyitself:
4 4 4 16Sowhathappenswhenyoutrytotakethesquarerootofsomething?Squaresandsquarerootsareinversefunctions,sooneβundoesβtheother.Mathematically,itwouldlooklikethis:
3 and 3 bothequal9 β¦whichmeansβ¦β9equals3
β9equals 3
β9equals 3Physically,itwouldlooklikethis:
9dotssplitupinto3rowsand3columns
Thesquarerootisdescribingthematchinglength&widthofthearrayβββthatbothsidesare3.Sowhathappensifitβsnotaperfectsquare,like9or16?Well,welookforaperfectsquareamongitsfactors.Forexample,32isnotaperfectsquare,butitequals16 2.Thenwedealwiththeperfectsquareportionofthenumber,andleavetherestalone.
Bytheendofthislesson,youshouldbeableto: identifywhenasquarerootsimplifiestoanintegerandwhenitdoesnβt
simplifythesquarerootofanonβperfectsquare solveaquadraticbytakingthesquarerootofanonβperfectsquare
Page 42
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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Hereβshowitlooksmathematically:
32 16 2 β32 β16 2 β16 β2Now,youmightbethinking,βHey,Ithoughtsquarerootscouldnβtdistributeovermultipleterms,βandyouwouldbecorrect.Butremember,atermisacollectionofthingsbeingmultiplied,andthatβswhatthe16and2aredoing.Sowecantakeonesquarerootoftwonumbersmultiplied,andrewriteitasonemultiplicationoftwosquarerootsofnumbers.Oncewegettotheβonemultiplicationoftwosquarerootsβ,whathappensnextisdifferentforthepartthatisaperfectsquarethanforthepartthatisnβt:
β16 β2
4 β2
whichweusuallywriteas: 4β2Whenwetrytorepresentthisprocessusinganarray,itlookssomethinglikethis:
32dotsin2separate4by4arrays
Sointheanswer4β2,the4indicatesthesidelengthofthearray,andthe2isthefactthatweneedtwooftheminordertoholdatotalof32dots.Letβsseethatprocessafewmoretimes,withdifferentnumbers:
75 25 3andso
β75 β25 β3
5 β3
whichwewriteas:5β3
18 9 2andso
β18 β9 β2
3 β2
whichwewriteas:3β2
80 16 5andso
β80 β16 β5
4 β5
whichwewriteas:4β5
600 100 6andso
β600 β100 β6
10 β6
whichwewriteas:10β6
Page 43
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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1. Copyandcompletethefollowingsimplifications:
a. 98 49 2 β98 β49 β ______β_____
b. 48 16 3 β48 β β3 ______β_____
Example2 So,howdoweknowwhattwofactorstouseinordertoexpressanonβperfectsquareasaproductofaperfectsquareandsomethingelse?Mostly,byguessingandchecking.
Letβssaythatyouwantedtosimplify:β128Thelistofthefirst12perfectsquaresis: 1 4 9 16 25 36 49 64 81 100 121 144For128,wecouldstartwiththenextsmallestperfectsquare,andworkourwaydown:
Does121gointo128?NO Does100gointo128?NO Does81?NO Does64?YESβββ64 2
SonowwecancontinuewiththeprocessusedinExample1:
128 64 2 β128 β64 β2
8 β2
whichwewriteas: 8β2Whenusingthisprocess,however,wemustbecarefultostartatthetopofthelistandworkourwaydown.Ifwedonβtusethelargestperfectsquarethatdividesthenumber,thenweβreleavingmoreperfectsquaresβtrappedβinsidethesquareroot.Forexample,128canalsobedividedby16,likethis:
β128 β16 β8
8β8Anditcanalsobedividedby4,likethis:
β128 β4 β32
2β32
But β8 can be divided by
another perfect square: 4
But β32 can be divided by
another perfect square: 16
Page 43
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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Moralofthestory:findthelargestperfectsquarethatdividesintothenumberyouβretryingtosimplify.Itisalsobestifyouputtheperfectsquarefirst,thentheotherfactor,becauseitβstheperfectsquarethatwillgeneratethecoefficientontheoutsideofthesquarerootsymbol.Ifwemissfindingthelargestperfectsquare,wecanβpiggybackβoffofwhatwedofind.Takethatlast
versionofβ128,forexample:
β128 β4 β32
2β32
2β16 β2
2 4 β2
Forafinalanswerof: 8β2CAUTION:Noteveryradicand(thatβsthenumberunderthesquarerootsymbol)canbesimplified.
Forexample,β10isnotsimplifiablebecausenoneofthefactorsof10(2&5)areperfectsquares.Ifnoperfectsquaredividesthenumberevenly,thenthesquarerootstaysasis,nochanges.
2. ThesimplificationsstartedbelowdoNOTusethelargestperfectsquarepossible.Findthelargestperfectsquarethatdividestheradicandandcompletethesimplification.
a. β200 β25 β8 b. β72 β4 β18
3. Eachofthesimplificationsbelowhasbeendoneincorrectly.Describeandfixthemistake.
a. β150 β50 β3 25β3
b. β120 β30 β4 15β2
4. Describewhatiswrongwiththesetwosimplifications:
a. β15 3β5 b. β36 6β6
Example3 Theprevious2examplesusedtheideaofbreakingaradicandintoaperfectsquarefactorandanonβperfectβsquarefactor.Butthereisanotherwaytovisualizetheideaofsimplifyingasquareroot,anditgoesbacktoamethodyoustudiedinpreviousmathclasses:primefactorization.Theideabehindprimefactorizationisthatinsteadofbreakinganumberintoonlyapairoffactors,weinsteadbreakitdowntoalistoffactorsthatareallprimenumbers.Forexample,insteadofthinkingof18as9 1,wevisualizeitas3 3 2.
But β32 can be divided by
another perfect square: 16
Page 44
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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Howdoesthisrelatetosimplifyingaperfectsquare?Well,rememberthatsquaringsomethingmeansmultiplyingitbyitself.Sosince18 3 3 2thenwecanwriteitas18 3 2.Andifasquarerootβundoesβasquare,then:
18 3 2 β18 3 β2Thenthesquarerootcancelsthepower,likethis:
β18 3 β2
Leavinguswiththis: β18 3β2Letβsseethatprocessagain,withsomedifferentnumbers:
50 5 10whichisalso5 5 2andthatmeans
β50 5 β2
5 β2
whichwewriteas:5β2
56 8 7whichisalso2 2 2 7
andthatmeans
β56 2 β14
2 β14
whichwewriteas:2β14
ThisprocesscanalsoprovetousthatsomethingisNOTabletobesimplified:
30 6 5whichisalso2 3 5
andsincenoneofthefactorsarerepeated,thereisnowaytosimplifythesquareroot
210 21 10whichisalso3 7 2 5
andsincenoneofthefactorsarerepeated,thereisnowaytosimplifythesquareroot
CAUTION:Thereisadifferencebetweensimplifyingasquarerootandestimatingasquareroot.Tosimplifyasquarerootmeanstoremoveanyfactorsthatrepresentperfectsquares.Toestimateasquarerootmeanstogiveadecimalapproximationofthevalue.
Forexample:β45simplifiesto3β5becauseof45isamultipleof9(specifically,9 5).Buttheestimateofitsvalueis6.708β¦because45isbetween36(whichis6 )and49(whichis7 ).
5. Usetheprimefactorizationsshowntosimplifythesquareroots:
a. 96 2 2 19andsoβ96 _____β_____
b. 126 2 3 3 7andsoβ126 _____β_____
6. Useprimefactorizationstoshowwhythefollowingsquarerootscanβtbesimplified:
a. β66
b. β70
Page 45
NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I
Lesson 9: Simplifying NonβPerfect Square Roots Last edited on 7/19/17
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ProblemSetSimplifyeachsquareroot.
1. β24
2. β40
3. β54
4. β90
5. β80
6. β112
7. β98
8. β108
9. β192
10. β28
11. β8
12. β125
13. β216
14. β27
15. β300
16. β288
17. β175
18. β72
19. β44
20. β245
Describe/explainthemistakemadeineachsimplificationshownbelow:
21. β14 2β7
22. β20 5β4
23. β72 3β8
24. β25 β5
Page 46
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 10 ALGEBRA I
Lesson 10: More Solving by Taking the Square Root Last edited on 7/22/17
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Lesson 10: More Solving by Taking the Square Root
Example 1
Lesson 8 discussed how to solve equations by taking the square root of both sides. In that lesson, we
only dealt with the square roots of perfect squares. Lesson 9 looked at how to take the square roots of
non-perfect squares. This lesson puts those two ideas together.
Here is how the process looked in Lesson 8:
π₯2 = 25
Β±βπ₯2 = Β±β25
π₯ = Β±5
π₯2 β 49 = 0 +ππ + ππ
π₯2 = 49
Β±βπ₯2 = Β±β49
π₯ = Β±7
3π₯2 = 12 Γ· π Γ· π
π₯2 = 4
π₯ = Β±2
2π₯2 + 10 = 28 βππ β ππ
2π₯2 = 18 Γ· π Γ· π
π₯2 = 9
π₯ = Β±3
In each of these examples, we used the Properties of Equality to isolate the π₯2 term and then took the
square root of both sides. If we apply this solution process on equations involving non-perfect
squares, the solution process stays the same but answers look different. Compare these examples:
Perfect Square Non-Perfect Square
π₯2 = 25
Β±βπ₯2 = Β±β25
π₯ = Β±5
π₯2 = 24
Β±βπ₯2 = Β±β24
π₯ = Β±β4 Γ β6
π₯ = Β±2β6
By the end of this lesson, you should be able to:
solve a quadratic equation by taking the square root
identify when taking the square root will have integer answers and when it wonβt
present non-integer answers as conjugate pairs of rational and irrational numbers
when needed
Page 47
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 10 ALGEBRA I
Lesson 10: More Solving by Taking the Square Root Last edited on 7/22/17
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Perfect Square Non-Perfect Square
π₯2 β 49 = 0 +ππ + ππ
π₯2 = 49
Β±βπ₯2 = Β±β49
π₯ = Β±7
π₯2 β 48 = 0 +ππ + ππ
π₯2 = 48
Β±βπ₯2 = Β±β48
π₯ = Β±β16 Γ β3
π = Β±πβπ
3π₯2 = 12 Γ· π Γ· π
π₯2 = 4
π₯ = Β±2
4π₯2 = 12 Γ· 4 Γ· 4
π₯2 = 3
π = Β±β3
2π₯2 + 10 = 28 βππ β ππ
2π₯2 = 18 Γ· π Γ· π
π₯2 = 9
π = Β±π
2π₯2 + 10 = 46 βππ β ππ
2π₯2 = 36 Γ· π Γ· π
π₯2 = 18
π₯ = Β±β18 πππππππ β Β± β9 Γ β2
π₯ = Β±3β2
As you can see, the Properties of Equality are still applied, as usual. But instead of the square roots
providing integer answers, we get instead a radicand that simplifies (but stays irrational).
1. Copy and solve the following in your notes:
a. 5π₯2 β 40 = 0 b. 2π₯2 + 5 = 95
(You should get Β±2β2 for part a and Β±3β5 for part b.)
Example 2
The other type of problem discussed in Lesson 8 was this style:
(π₯ + 3)2 = 16
Β±β(π₯ + 3)2 = Β±β16
π₯ + 3 = Β±4 βπ β π
π₯ = β3 Β± 4
π₯ = β3 + 4 ππ π₯ = β3 β 4
π₯ = 1 ππ π₯ = β7
What makes this type of problem distinct from the ones in Example 1 is the fact that the quadratic is a
binomial squared instead of a monomial squared (in other words, the fact that it has parentheses).
Page 48
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 10 ALGEBRA I
Lesson 10: More Solving by Taking the Square Root Last edited on 7/22/17
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So, what happens if the binomial squared is not isolated on one side of the equation? Well, our first job
will be to isolate it. And for that, we continue to use the trusty Properties of Equality.
Consider this problem, for example: 2(π₯ + 3)2 β 40 = 0
We first get rid of the 40: 2(π₯ + 3)2 = 40
And then get rid of the 2: (π₯ + 3)2 = 20
At this point, we square root both sides: Β±β(π₯ + 3)2 = Β±β20
The root and power cancel: π₯ + 3 = Β±β20
And we simplify the radical: π₯ + 3 = Β±β4 Γ β5
π₯ + 3 = Β±2β5
The next thing is to get rid of the 3: βπ β π
But at this point, there is a problem. The β3 is rational. The mixed radical 2β5 is irrational ---
a non-repeating, non-terminating decimal. So they are not like terms, and we canβt combine them.
We could add β3β5 with 2β5, or we could add β3 with 2, but we canβt mix a regular number
with a radical. (We could estimate their total approximate decimal value, but we canβt combine them.)
So what does this mean for the final answer? Well, it doesnβt go much farther:
π₯ = β3 Β± 2β5
The βplus-or-minusβ indicates that what appears to be a single answer is actually a conjugate pair of
answers, like this:
π₯ = β3 + 2β5 or π₯ = β3 β 2β5
2. Copy and solve each equation in your notes. Show all steps.
a. 4(π₯ β 1)2 β 32 = 0 b. 3(π₯ + 2)2 + 3 = 45
(You should get 1 Β± 2β2 for part a and β2 Β± β14 for part b.)
Example 3
As you saw in Lesson 8, sometimes a square root reduces to a whole number. It depends on whether
the radicand is a perfect square or not.
When solving an equation with a binomial square, like in Example 2, if the radicand is a perfect square,
then the answers will be two integers. Instead of leaving the answer as a conjugate pair of rational and
irrational numbers mixed together, we will instead continue on for one more step.
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 10 ALGEBRA I
Lesson 10: More Solving by Taking the Square Root Last edited on 7/22/17
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Hereβs how that would look:
3(π₯ β 5)2 + 2 = 20 β2 β 2
3(π₯ β 5)2 = 18 Γ· 3 Γ· 3
(π₯ β 5)2 = 9
Β±β(π₯ β 5)2 = Β±β9
π₯ β 5 = Β±3 +5 + 5
π₯ = 5 Β± 3
π₯ = 5 + 3 ππ π₯ = 5 β 3
π₯ = 8 ππ π₯ = 2
3. Copy and solve each equation in your notes. Show all steps.
a. 6(π₯ β 4)2 β 1 = 5 b. β3(π₯ + 2)2 + 8 = β40
(One of the answers to part a is 5, and one of the answers to part b is β6.)
Problem Set
Solve each equation.
1. π₯2 + 1 = 81
2. 12π₯2 = 144
3. 3π₯2 β 8 = 10
4. 4π₯2 β 11 = 49
5. 2(π₯ β 6)2 + 10 = 98
6. 3(π₯ + 8)2 β 12 = 48
7. 3(π₯ + 2)2 β 9 = 36
8. 5(π₯ β 7)2 + 1 = 126
9. 2(π₯ + 3)2 + 2 = 100
10. 10(π₯ β 5)2 + 6 = 46
partial answers: 1) 4β5 3) β6 5) 6 + 2β11 7) β2 + β15 9) β10
Page 50
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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Lesson 11: Completing the Square
Example 1 In Module 3, we spent some time discussing parent functions. One of the functions we studied was ππ(π₯π₯) = π₯π₯2. You learned how changes to the function altered the shape and placement of the graph. For example, given the function ππ(π₯π₯) = (π₯π₯ β 3)2 β 5, we know that the parent graph has been shifted to the right 3 and down 5. This means that the vertex of the function is the point (3,β5). The format ππ(π₯π₯) = (π₯π₯ β β)2 + ππ is known as the vertex form of a quadratic because it shows the vertex coordinates. Consider the examples below:
function vertex
ππ(π₯π₯) = (π₯π₯ + 4)2 + 7 (β4, 7)
ππ(π₯π₯) = (π₯π₯ β 2)2 + 3 (2, 3)
ππ(π₯π₯) = β(π₯π₯ + 5)2 (β5, 0)
ππ(π₯π₯) = 12
(π₯π₯ β 1)2 β 8 (1,β8)
In this lesson, we are going to discuss how to take a function that is presented in standard form, where the terms are in order by highest to lowest power, and transform it into vertex form. This process will take advantage of the pattern that was reviewed in Lesson 4 regarding binomial squares. Here is that pattern again:
(4π₯π₯ β 7)2 ππππππππππππππ 16π₯π₯2 β 56π₯π₯ + 49 But we are going to focus on problems where the leading coefficient is just a 1. That means the pattern is a little simpler:
(π₯π₯ β 7)2 ππππππππππππππ π₯π₯2 β 14π₯π₯ + 49
By the end of this lesson, you should be able to: β’ convert a quadratic function from standard form (with a leading coefficient of 1)
into vertex form by completing the square β’ solve a quadratic function by completing the square
4π₯π₯, squared β7, squared 4π₯π₯ times β7 times 2
π₯π₯, squared β7, squared π₯π₯ times β7 times 2
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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And then that pattern was able to be worked backwards in order to factor a quadratic back to a binomial square. For example:
Given the problem: π₯π₯2 β 8π₯π₯ + 16 This problem factors as:
(π₯π₯ β 4)2 What weβre going to be doing is forcing every quadratic we come across to fit that binomial square pattern. This is known as completing the square. Letβs say that we were given the quadratic: ππ(π₯π₯) = π₯π₯2 β 8π₯π₯ + 17 Thatβs pretty close to the example at the top of the page: π₯π₯2 β 8π₯π₯ + 16 The constant term is just off by 1, at 17 instead of 16. So what if we separated out that extra one? What if we rewrote the problem like this:
ππ(π₯π₯) = π₯π₯2 β 8π₯π₯ + 16οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βπππ‘π‘ π‘π‘βππππ ππππ
(π₯π₯β4)2
+ 1
Then we can write the function as: ππ(π₯π₯) = (π₯π₯ β 4)2 + 1 which is in vertex form. Letβs see that again:
binomial square: completing the square:
ππ(π₯π₯) = π₯π₯2 β 10π₯π₯ + 25 becomes
ππ(π₯π₯) = (π₯π₯ β 5)2
ππ(π₯π₯) = π₯π₯2 β 10π₯π₯ + 27 becomes
ππ(π₯π₯) = π₯π₯2 β 10π₯π₯ + 25οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
(π₯π₯β5)2
+ 2
so the final version is: ππ(π₯π₯) = (π₯π₯ β 5)2 + 2
ππ(π₯π₯) = π₯π₯2 + 14π₯π₯ + 49 becomes
ππ(π₯π₯) = (π₯π₯ + 7)2
ππ(π₯π₯) = π₯π₯2 + 14π₯π₯ + 54 becomes
ππ(π₯π₯) = π₯π₯2 + 14π₯π₯ + 49οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
(π₯π₯β5)2
+ 5
so the final version is: ππ(π₯π₯) = (π₯π₯ + 7)2 + 5
π₯π₯ squared β4 squared π₯π₯ times β4 times 2
π₯π₯ squared β5 squared π₯π₯ times β5 times 2
π₯π₯ squared 7 squared π₯π₯ times 7 times 2
Page 52
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 9 becomes
ππ(π₯π₯) = (π₯π₯ β 3)2
ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 8 becomes
ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 9οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
(π₯π₯β3)2
β 1
so the final version is: ππ(π₯π₯) = (π₯π₯ β 3)2 β 1
ππ(π₯π₯) = π₯π₯2 + 20π₯π₯ + 100 becomes
ππ(π₯π₯) = (π₯π₯ + 10)2
ππ(π₯π₯) = π₯π₯2 + 20π₯π₯ + 90 becomes
ππ(π₯π₯) = π₯π₯2 + 20π₯π₯ + 100οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
+102
β 10
so the final version is: ππ(π₯π₯) = (π₯π₯ + 10)2 β 10
1. Copy and complete the table in your notes:
binomial square: completing the square:
ππ(π₯π₯) = π₯π₯2 β 12π₯π₯ + 36 becomes
ππ(π₯π₯) = (π₯π₯ β )2
ππ(π₯π₯) = π₯π₯2 β 12π₯π₯ + 38 becomes
ππ(π₯π₯) = π₯π₯2 β 12π₯π₯ + _____οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
(π₯π₯β )2
+ _____
so the final version is: ππ(π₯π₯) = (π₯π₯ β ______)2 + ______
ππ(π₯π₯) = π₯π₯2 + 16π₯π₯ + 64 becomes
ππ(π₯π₯) = (π₯π₯ + )2
ππ(π₯π₯) = π₯π₯2 + 16π₯π₯ + 54 becomes
ππ(π₯π₯) = π₯π₯2 + ____ π₯π₯ + _____οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π€π€π€π€ πππππππ€π€ π‘π‘βππππ ππππ
(π₯π₯+ )2
β _____
so the final version is: ππ(π₯π₯) = (π₯π₯ + ______)2 β ______
Example 2 So far, weβve relied on being able to recognize a pattern and use it to our advantage. And thatβs a pretty good strategy for a lot of problems. But we want to develop this process into a series of steps that will always work, regardless of our ability to recognize an underlying pattern.
π₯π₯ squared β3 squared π₯π₯ times β3 times 2
π₯π₯ squared 10 squared π₯π₯ times 10 times 2
π₯π₯ squared _____ squared π₯π₯ times _____ times 2
π₯π₯ squared _____ squared π₯π₯ times β6 times 2
Page 53
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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Step 1 is to clear the work space a littleβ¦ having the constant term hovering around, and trying to visualize how it connects to the square youβre trying to make, is distracting. So we will use the Additive Property of Equality to put the constant on the other side of the equals sign, like this:
ππ(π₯π₯) = π₯π₯2 + 12π₯π₯ + 8 βππ β ππ
ππ(π₯π₯) β 8 = π₯π₯2 + 12π₯π₯ Step 2 is to recognize that the linear term (with the π₯π₯) is always a multiple of 2. We need to see it without that doubling, so we divide the linear term by 2. However, we do NOT divide on both sides. The purpose of dividing is to find what goes in with the π₯π₯ in the binomial square. So it looks like this:
ππ(π₯π₯) β 8 = π₯π₯2 + 12π₯π₯ Γ· 2
ππ(π₯π₯) β 8 = (π₯π₯ + 6) Step 3 is to remember that what weβre trying to create is a binomial square. So the next thing to do is to write a power of 2 on the outside of the binomial, like this:
ππ(π₯π₯) β 8 = (π₯π₯ + 6)2 Once weβve written in that power, though, thereβs an issue --- weβre guaranteeing that the 6 is squared. That adds 36 to the right hand side. (You canβt see the 36 --- itβs hidden as part of the binomial square.) Step 4 is to balance that extra constant by adding the squared constant to the other side:
ππ(π₯π₯) β 8 = (π₯π₯ + 6)2 +ππππ
Now we have this: ππ(π₯π₯) + 28 = (π₯π₯ + 6)2 Step 5, the last step, is to put the constant back on the right, to serve as part of the vertex:
ππ(π₯π₯) + 28 = (π₯π₯ + 6)2 βππππ βππππ
For a final answer of: ππ(π₯π₯) = (π₯π₯ + 6)2 β 28 Hereβs another example, without the discussion in between each step:
Step 1: move the constant to the left ππ(π₯π₯) = π₯π₯2 β 10π₯π₯ β 7 +ππ +ππ
Step 2: divide linear term by 2; put the # in parentheses with the π₯π₯ Step 3: write a power of 2 on the parentheses
ππ(π₯π₯) + 7 = π₯π₯2 β 10π₯π₯ Γ· 2
ππ(π₯π₯) + 7 = (π₯π₯ β 5)2 Step 4: add the constant squared to the left side +ππππ
ππ(π₯π₯) + 32 = (π₯π₯ β 5)2 Step 5: put the constant back on the right βππππ βππππ
ππ(π₯π₯) = (π₯π₯ β 5)2 β 28
Note: we do NOT need to add 36 to both sides --- writing in the power already added 36 to the right hand side
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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And by the way, it wouldnβt matter if the problem presented the quadratic using ππ(π₯π₯) = notation, or if it used π¦π¦ = notation. The process would stay the same:
Step 1: move the constant to the left π¦π¦ = π₯π₯2 + 6π₯π₯ + 15 βππππ βππππ
Step 2: divide linear term by 2; put the # in parentheses with the π₯π₯ Step 3: write a power of 2 on the parentheses
π¦π¦ β 15 = π₯π₯2 + 6π₯π₯ Γ· 2
π¦π¦ β 15 = (π₯π₯ + 3)2 Step 4: add the constant squared to the left side +ππ
π¦π¦ β 6 = (π₯π₯ + 3)2 Step 5: put the constant back on the right +ππ + ππ
π¦π¦ = (π₯π₯ + 3)2 + 6 2. Copy each equation into your notes. Complete the square to convert them into vertex form. Show
all steps. a. ππ(ππ) = ππππ β ππππ β ππ b. ππ = ππππ + ππππ + ππ
Example 3 The completing the square process can be used to solve quadratic equations as well as to convert quadratic functions into vertex form. The problems wil not be presented in the same way, however --- instead of the left hand side being either ππ(ππ) = or ππ = , the left side will have the quadratic, and the right hand side will have either zero or a constant. Here is an example:
ππππ + ππππππ β ππ = βππππ The complete the square process will be operating on the part of the equation that has the ππβs, so we get the 8 to join the βππππ, and then ignore that subtotal for a little while:
ππππ + ππππππ β ππ = βππππ +ππ +ππ ππππ + ππππππ = βππ
Γ· ππ (ππ + ππ)ππ = βππ +ππππ (ππ + ππ)ππ = ππππ
At this point, instead of bringing the 45 back to the same side as the binomial square, we solve the problem by taking the square root of both sides, like we did in Lesson 10:
οΏ½(ππ + ππ)ππ = Β±βππππ
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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And in this case, the radicand is not a perfect square, so we end up simplifying to a mixed radical: οΏ½(ππ + ππ)ππ = Β±βππππ
Β±βππ Γ βππ ππ + ππ = Β±ππβππ
βππ βππ ππ = βππ Β± ππβππ
So the final answers are: ππ = βππ + ππβππ and ππ = βππ β ππβππ It is possible for the radicand to be a perfect square, and in that situation, the answers are two integers:
ππππ β ππππππ + ππππ = ππππππ βππππ β ππππ ππππ β ππππππ = ππππ
Γ· ππ (ππ β ππ)ππ = ππππ +ππππ (ππ β ππ)ππ = ππππππ
οΏ½(ππ β ππ)ππ = Β±βππππππ ππ β ππ = Β±ππππ +ππ + ππ ππ = ππ Β± ππππ
ππ = ππ + ππππ ππππ ππ = ππ β ππππ ππ = ππππ ππππ ππ = βππ
3. Copy and solve each equation in your notes. Show all steps.
a. π₯π₯2 β 4π₯π₯ + 7 = 15 b. π₯π₯2 + 8π₯π₯ β 5 = 28 (Part a has a mixed radical/rational answer and part b has two integer answers.)
Page 56
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 11 ALGEBRA I
Lesson 11: Completing the Square Last edited on 3/5/18
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Problem Set Solve each equation. 1. π₯π₯2 + 2π₯π₯ + 9 = 21
2. π₯π₯2 β 4π₯π₯ β 7 = 13
3. π₯π₯2 β 8π₯π₯ + 20 = 40
4. π₯π₯2 + 10π₯π₯ β 4 = 35
Convert each function to vertex form: 5. ππ(π₯π₯) = π₯π₯2 + 6π₯π₯ β 4
6. ππ(π₯π₯) = π₯π₯2 β 12π₯π₯ + 40
7. π¦π¦ = π₯π₯2 + 14π₯π₯ + 36
8. π¦π¦ = π₯π₯2 β 16π₯π₯ + 50
partial answers: 1) β1 + β13 2) 2 + 2β6 3) β2 4) β13
Page 57
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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Lesson 12: Completing the Square with ππ β ππ
Example 1 In the last lesson, we discussed how to use a process known as completing the square. Itβs one way to convert a quadratic function into vertex form, and itβs also one way to solve a quadratic equation. You have learned, and will learn in the future, other ways to do both of those things. For example, letβs say you wanted to solve the quadratic equation: π₯π₯2 β 6π₯π₯ + 5 = 0 We already know that we canβt just subtract the 5 and divide by 6, for example, nor can we square root to get rid of the power. Whenever there is a mixture of powers, we have to either factor or use the complete the square process. Here is the solution process for the equation, done with both methods:
factoring: completing the square:
π₯π₯2 β 6π₯π₯ + 5 = 0 think: what multiplies to 5 and adds to β6?
(π₯π₯ β 5)(π₯π₯ β 1) = 0 π₯π₯ β 5 = 0 ππππ π₯π₯ β 1 = 0 +ππ + ππ +ππ + ππ π₯π₯ = 5 ππππ π₯π₯ = 1
ππππ β ππππ + ππ = ππ βππ β ππ ππππ β ππππ = βππ
Γ· ππ (ππ β ππ)ππ = βππ +ππ (ππ β ππ)ππ = ππ
οΏ½(ππ β ππ)ππ = Β±βππ ππ β ππ = Β±ππ +ππ + ππ ππ = ππ Β± ππ
ππ = ππ + ππ ππππ ππ = ππ β ππ ππ = ππ ππππ ππ = ππ
Most people would agree that the completing the square process looks a little longer, but keep in mind that many of the steps shown here would actually be done using mental math, and not written down. Obviously, if the quadratic is not factorable, then the factoring method wonβt work. For example, π₯π₯2 β6π₯π₯ + 4 = 0 is not factorable, but it could still be solved using the completing the square process. (The answers would not be integers, however.)
By the end of this lesson, you should be able to: β’ convert a quadratic function from standard form (with a leading coefficient other
than 1) into vertex form by completing the square β’ solve a quadratic with a leading coefficient other than 1 by completing the square
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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Up until now, we have only tried to complete the square with quadratics whose middle term was an even number. But what would happen if it was an odd number? Letβs say that we had the equation: π₯π₯2 β 5π₯π₯ + 6 = 0. This factors as: (π₯π₯ β 2)(π₯π₯ β 3) = 0 so the answers are 2 and 3. Letβs try completing the square and see if we get the same thing:
π₯π₯2 β 5π₯π₯ + 6 = 0 β6 β 6
π₯π₯2 β 5π₯π₯ = β6 Γ· 2
οΏ½π₯π₯ β 52οΏ½2
= β6
+ 254
οΏ½π₯π₯ β 52οΏ½2
= 14
οΏ½οΏ½π₯π₯ β 52οΏ½2
= Β±οΏ½ 1 4
π₯π₯ β 52
= Β± 12
+ 52
+ 52
π₯π₯ = 5 2
Β± 1 2
π₯π₯ = 52
+ 12
ππππ π₯π₯ = 52β 1
2
π₯π₯ = 62
ππππ π₯π₯ = 42
π₯π₯ = 3 ππππ π₯π₯ = 2 As we can see from this example, when the problem has an odd middle term, completing the square still works, but we will have to cope with fractions.
1. Use the completing the square process to solve the following quadratic. As you work, explain how the fractions squared, and how you took the square root of them. Also show how you converted to common denominators and how you simplified to lowest terms.
π₯π₯2 β 9π₯π₯ β 10 = 0 Because this problem is factorable, you should get integer answers --- specifically:
π₯π₯ = 10 ππππ π₯π₯ = β1 Be aware that if the problem isnβt factorable, then completing the square will lead to conjugate pairs of mixed fractions & radicals.
When squaring a fraction, the top and the bottom are both squared:
οΏ½ 5 2οΏ½2
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ (5)2
(2)2
β6 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β24
4
and then β24 4
+ 254 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 1
4
When taking the square root of a fraction, the root applies to both top & bottom:
Β±οΏ½ 1 4 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½Β± β1
β4
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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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Here is an unfactorable, odd-middle-term quadratic equation, solved by completing the square: π₯π₯2 + 3π₯π₯ β 7 = 0
+ππ + ππ π₯π₯2 + 3π₯π₯ = 7
Γ· 2
οΏ½π₯π₯ + 32οΏ½2
= 7
+ ππππ
οΏ½π₯π₯ + 32οΏ½2
= 374
οΏ½οΏ½π₯π₯ + 32οΏ½2
= Β±οΏ½ 37 4
π₯π₯ + 32
= Β± β372
βππππ
β ππππ
π₯π₯ = β 3 2
Β± β37 2
π₯π₯ = β32
+ β372
ππππ π₯π₯ = β32β β37
2
This particular radicand does not simplify, but even if it could, we wouldnβt be able to combine it with the β3. We could write them as single fractions, like this:
π₯π₯ = β3 + β372
ππππ π₯π₯ = β3 β β372
but that doesnβt really make it any nicer, so itβs not strictly necessary.
1. Use the complete the square process to solve these non-factorable quadratics: a. π₯π₯2 β 7π₯π₯ + 4 = 0 b. π₯π₯2 + π₯π₯ β 3 = 0
Example 2 Okay, weβve figured out how to cope when the middle number isnβt the perfect thing to divide by 2. Now itβs time to figure out how to cope when the first number isnβt the perfect thing, either. So far, weβve focused on quadratics whose leading coefficient is 1, because itβs easier to complete the square with them. That makes learning the process simpler. But now that weβve seen the process a few times, letβs see what happens when the leading coefficient is something other than 1. The first step of completing the square is to move the constant. Itβs a way to clear the decks a little bit and make room to maneuver. When we have a leading coefficient that isnβt 1, then part of that βclearing the decksβ process is to get rid of that coefficient.
Page 58 Page 60
Β±οΏ½37 4 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ Β± β37
β4
οΏ½32οΏ½2
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ (3)2
(2)2
7 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 28
4
and then 28 4
+ 94 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 37
4
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Hereβs an example: Step 1: move the constant
AND 3π₯π₯2 + 30π₯π₯ + 15 = 0
βππππ βππππ divide away the coefficient
Step 2: divide linear term by 2; put the # in parentheses with the π₯π₯
Step 3: write a power of 2 on the parentheses
3π₯π₯2 + 30π₯π₯ = β15 Γ· ππ Γ· ππ Γ· ππ
π₯π₯2 + 10π₯π₯ = β5 Γ· ππ
(π₯π₯ + 5)2 = β5 Step 4: add the constant squared to the other side +ππππ
(π₯π₯ + 5)2 = 20 Step 5: take the square root; simplify as needed Step 6: finish solving for π₯π₯
οΏ½(π₯π₯ + 5)2 = Β±β20 π₯π₯ + 5 = Β±β4 Γ β5
π₯π₯ + 5 = Β±2β5 βππ βππ
π₯π₯ = β5 Β± 2β5
2. Copy each equation into your notes. Complete the square to solve. Show all steps. a. 2π₯π₯2 β 12π₯π₯ β 8 = 0 b. 5π₯π₯2 + 10π₯π₯ + 40 = 0
Example 3 Examples 1 & 2 used completing the square in order to solve quadratic equations. But the same process can be used to convert quadratic functions into vertex form. Here is an example with an odd middle term:
ππ(ππ) = ππππ + ππππππ+ ππ βππ βππ ππ(ππ) β ππ = ππππ + ππππππ
Γ· ππ
ππ(ππ) β ππ = οΏ½ππ + πππππποΏ½ππ
+ ππππππππ
ππ(ππ) β ππππππ
+ ππππππππ
= οΏ½ππ + πππππποΏ½ππ
ππ(ππ) + ππππππ
= οΏ½ππ + πππππποΏ½ππ
βππππππ
βππππππ
ππ(ππ) = οΏ½ππ + πππππποΏ½ππβ ππππ
ππ
The quadratic is now in vertex form, and shows the vertex to be the point οΏ½β ππππππ
,βπππππποΏ½.
Page 59 Page 61
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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Here is an example with a leading coefficient other than 1: ππ = ππππππ β ππππππ β ππππ
+ππππ +ππππ ππ + ππππ = ππππππ β ππππππ
Γ· ππ Γ· ππ Γ· ππ Γ· ππ
ππ ππ
+ ππ = ππππ β ππππ Γ· ππ
ππ ππ
+ ππ = (ππ β ππ)ππ +ππ
ππ ππ
+ ππππ = (ππ β ππ)ππ βππππ βππππ
ππ ππ
= (ππ β ππ)ππ β ππππ At this point, we multiply through by the 4 to get the ππ by itself:
πποΏ½πππποΏ½ = ππ(ππ β ππ)ππ β ππ(ππππ)
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ ππ = ππ(ππ β ππ)ππ β ππππ
Now this quadratic is in vertex form, showing a vertex of (ππ,βππππ) and having a stretch factor of 4.
3. Copy and solve each equation in your notes. Show all steps. a. π¦π¦ = 2π₯π₯2 β 4π₯π₯ + 7 b. π¦π¦ = π₯π₯2 + 7π₯π₯ β 5
It is possible for a quadratic to have both a leading coefficient other than one and an odd middle term:
ππ = ππππππ + ππππππ + ππππ βππππ βππππ
ππ β ππππ = ππππππ + ππππππ Γ· ππ Γ· ππ Γ· ππ Γ· ππ
ππ ππβ ππ = ππππ + ππππ
Γ· ππ ππ ππβ ππππ
ππ= οΏ½π₯π₯ β ππ
πποΏ½ππ
+ ππππππ
ππ ππ
+ ππππππ
= οΏ½ππ β πππποΏ½ππ
βππππππ
βππππππ
ππ ππ
= οΏ½π₯π₯ β πππποΏ½ππβ ππππ
ππ
ππ = πποΏ½ππ β πππποΏ½ππβ ππππ
ππ
Notice how dividing affects both items on
the left side
Notice how dividing affects both items on
the left side
The 6 is converted
into 244
in order to add
all terms are multiplied by 3
Page 62
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 12 ALGEBRA I
Lesson 12: Completing the Square with ππ β ππ Last edited on 3/5/18
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Problem Set Solve each equation.
1. π₯π₯2 + 5π₯π₯ β 9 = 0
2. π₯π₯2 β 9π₯π₯ + 7 = 0
3. 5π₯π₯2 β 10π₯π₯ + 20 = 0
4. 6π₯π₯2 + 24π₯π₯ β 42 = 0
Convert each function to vertex form:
5. ππ(π₯π₯) = π₯π₯2 + 3π₯π₯ + 4
6. ππ(π₯π₯) = π₯π₯2 β 9π₯π₯ β 1
7. π¦π¦ = 2π₯π₯2 + 12π₯π₯ + 36
8. π¦π¦ = 3π₯π₯2 β 12π₯π₯ β 9
Page 63
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 13 ALGEBRA I
Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 13: Deriving the Quadratic Formula
Example 1 In the last two lessons, we discussed how to complete the square, both to convert a quadratic function into vertex form, and also to solve a quadratic equation. Completing the square can also be used to generate a formula known as the Quadratic Formula. This formula enables us to solve any quadratic equation by substituting the values from the equation into locations in the formula, without needing to complete the square. To generate the formula, we need to apply the completing the square process to a generic equation --- one in which the usual numeric coefficients have been replaced with letters:
πππ₯π₯2 + πππ₯π₯ + ππ = 0 We already know how to complete the square on an equation with numeric coefficients. We will use that to help us understand the generic equation by putting the two kinds side-by-side:
with numeric coefficients: with generic coefficients:
first, we βclear the decksβ because we want the first term to be just ππππ :
3π₯π₯2 + 11π₯π₯ + 5 = 0 βππ βππ
3π₯π₯2 + 11π₯π₯ = β5 Γ· ππ Γ· ππ Γ· ππ
π₯π₯2 + 113π₯π₯ = β5
3
ππππππ + ππππ + ππ = ππ βππ β ππ ππππππ + ππππ = βππ
Γ· ππ Γ· ππ Γ· ππ
ππππ + ππππππ = β ππ
ππ
because the terms do not divide evenly by the leading coefficient, we end up with fractions
next, we divide the middle term by 2, but because it doesnβt divide evenly, and with the way that fractions work, the denominator gets larger:
π₯π₯2 + 113π₯π₯ = β5
3
Γ· ππ
οΏ½π₯π₯ + 116οΏ½2
ππππ + ππππππ = β ππ
ππ
Γ· ππ
οΏ½ππ + πππππποΏ½ππ
By the end of this lesson, you should be able to: β’ apply the completing the square process to a generic quadratic equation β’ connect the completing the square process to the Quadratic Formula β’ use the Quadratic Formula on simple quadratic equations
Page 64
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 13 ALGEBRA I
Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18
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When we write the power on the outside of the parentheses, we balance that constant squared by adding the it to the other side as well:
οΏ½π₯π₯ + 116οΏ½2
= β 5 3
+ππππππππππ
οΏ½ππ + πππππποΏ½ππ
= β ππππ
+ππππ
ππππππ
*remember, (ππππ)ππ
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ ππππππ
but now we need to convert to a common denominator in order to combine the fractions on the right side:
οΏ½π₯π₯ + 116οΏ½2
= β 5 3 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β60
36
+ππππππππππ
6136
οΏ½ππ + πππππποΏ½ππ
= β ππππ ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ βππππππ
ππππππ
+ππππ
ππππππ
οΏ½ππππβπππππποΏ½ππππππ
*because the letters are not like terms, the numerator does not simplify
the next step is to take the square root of both sides, remembering that with fractions, the root applies to both the numerator and denominator:
οΏ½ οΏ½π₯π₯ + 116οΏ½2
= Β± β61β36
π₯π₯ + 116
= Β± β616
οΏ½οΏ½ππ + πππππποΏ½ππ
= Β± οΏ½(ππππβππππππ)οΏ½ππππππ
ππ + ππππππ
= οΏ½(ππππβππππππ)ππππ
last but not least, we isolate the variable using the Additive Property of Equality:
x + 116
= Β± β616
βππππππ
β ππππππ
π₯π₯ =β11 Β± β61
6
ππ + ππππππ
= οΏ½(ππππβππππππ)ππππ
β ππππππ
β ππππππ
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
And that final result is the Quadratic Formula:
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
You can see that we usually drop the parentheses around the portion that is underneath the square root symbol.
Page 65
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 13 ALGEBRA I
Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 2 Okay, so we derived the Quadratic Formula. Now, how do we use it to solve a quadratic equation? Letβs start with problems whose answers will not involve simplifying a square root. Hereβs an example:
5π₯π₯2 + 7π₯π₯ β 2 = 0
Step 1: identify ππ, ππ, and ππ
5π₯π₯2 + 7π₯π₯ β 2 = 0
ππ = ππ ππ = ππ ππ = βππ
Step 2: calculate ππ2 and 4ππππ
ππ2 = 49
4ππππ
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½4(5)(β2)
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β40
Step 3: write out the Quadratic Formula π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
Step 4: substitute the values from steps 1 & 2 into the formula
π₯π₯ =β7 Β± β49 ββ40
2(5)
Step 5: combine the #βs under the square root symbol AND under the fraction bar π₯π₯ =
β7 Β± β89 10
You can see that because the radicand is prime, the square root canβt be simplified. This is the end of the active calculation phase of the problem. We could write the two answers as:
π₯π₯ =β7 + β89
10 ππππππ π₯π₯ =
β7 β β8910
but that would be as far as we would go in this situation.
1. Copy the equation into your notes. Then copy and complete the Quadratic Formula set-up shown underneath.
a. ππππππ + ππππ β ππ = ππ
ππ2 = 4ππππ =
π₯π₯ =β3 Β± β β
2( ) ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
b. ππππππ β ππππ + ππ = ππ
ππ2 = 4ππππ =
π₯π₯ =9 Β± β β
2( ) ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
Page 58 Page 59 Page 59 Page 66
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 13 ALGEBRA I
Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 3 There are a few unusual things to watch out for when using the Quadratic Formula. One thing to realize is that if the middle term (the βbβ value) is negative, the Quadratic Formula will turn it back into a positive, like this:
2π₯π₯2 β 5π₯π₯ + 1 = 0 In this problem, = β5 , which would mean that ππ2 = +25 (not negative 25). When we apply the Quadratic Formula, it would look like this:
π₯π₯ =+5 Β± β25 β 8
2(2)
And the problem would proceed as usual. Another thing to realize is that if the first term (the βaβvalue) is negative, the Quadratic Formula will have a negative in the denominator, like this:
β3π₯π₯2 + 9π₯π₯ + 5 = 0 In this problem, ππ = β3, which will mean that 4ππππ = 4(β3)(5)
π€π€βππππβ ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β60
When we apply the Quadratic Formula, it would look like this:
π₯π₯ =β9 Β± β81 ββ60
2(β3)
As the formula calculations proceed, we end up here:
π₯π₯ =β9 Β± β141
β6
And technically speaking, weβre not supposed to leave a negative number in a denominator. (Remember, the denominator of a fraction is supposed to show how many pieces you cut a whole unit into, and itβs not possible to cut something into a negative number of pieces.) So we multiply both the top and bottom by β1:
π₯π₯ =οΏ½β9 Β± β141 οΏ½ Γ β1
(β6) Γ β1
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
9 Β± β1416
You can see that the sign change does not affect the radical --- it was already βplus-or-minusβ. One last thing to watch for is to make sure that the quadratic is set up with all the like terms brought onto the same side and combined:
4π₯π₯2 + 10π₯π₯ = 3π₯π₯ β 2 βππππ + ππ βππππ + ππ
4π₯π₯2 + 7π₯π₯ + 2 = 0 And then the Quadratic Formula can be applied:
π₯π₯ =β7 Β± β49 β 32
2(4) ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
β7 Β± β178
Page 58 Page 67
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 13 ALGEBRA I
Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Problem Set Use the Quadratic Formula:
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ with πππ₯π₯2 + πππ₯π₯ + ππ = 0
to solve each equation.
1. 3π₯π₯2 + 5π₯π₯ β 1 = 0
2. 2π₯π₯2 β 11π₯π₯ + 7 = 0
3. 5π₯π₯2 β π₯π₯ β 3 = 0
4. π₯π₯2 + 3π₯π₯ β 2 = 0
5. 4π₯π₯2 + 7π₯π₯ + 2 = 0
6. β7π₯π₯2 β 9π₯π₯ + 1 = 0
Page 62 Page 68
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 14 ALGEBRA I
Lesson 14: Using the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 14: Using the Quadratic Formula
Example 1 In the last lesson, we introduced how to use the Quadratic Formula, but we limited ourselves to equations whose answers involved non-reducible radicals. In this lesson, weβll expand the use of the Quadratic Formula to equations whose answers involve radicands that simplify. For example:
5π₯π₯2 + 9π₯π₯ + 2 = 0 will have a radical that does not simplify
5π₯π₯2 + 8π₯π₯ + 2 = 0 will have a radical that simplifies
We already know how to use the Quadratic Formula. Letβs see the differences between these two equations:
ππππππ + ππππ + ππ = ππ ππππππ + ππππ + ππ = ππ
first, we identify a, b, and c, then calculate ππππ and ππππππ :
ππ = 5, ππ = 9, ππ = 2
ππ2 = 81, 4ππππ = 4(5)(2) = 40
ππ = ππ, ππ = ππ, ππ = ππ
ππππ = ππππ, ππππππ = ππ(ππ)(ππ) = ππππ
next, we plug into the Quadratic Formula:
π₯π₯ =β9 Β± β81 β 40
2(5) π₯π₯ =βππ Β± βππππ β 40
2(5)
now we combine inside the radical, and in the denominator:
π₯π₯ =β9 Β± β41
10 π₯π₯ =
β8 Β± β44 10
in the left hand column, the radicand is prime, and does not simplify. in the right hand column, the radicand is a multiple of 4, and so it will simplify:
π₯π₯ =β9 Β± β41
10 π₯π₯ =
β8 Β± β4 Γ β1110 ππππππππππππππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β8 Β± 2β11
10
This is where the calculations end --- we canβt combine integers with mixed radicals. The only other
By the end of this lesson, you should be able to: β’ use the Quadratic Formula on any quadratic equation β’ use the discriminant to predict the number of solutions for a quadratic equation
Page 69
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 14 ALGEBRA I
Lesson 14: Using the Quadratic Formula Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
thing we could do to change the appearance of the answers at this point would be to write them as separate conjugate pairs of fractions, like this:
π₯π₯ =β9 Β± β41
10 π₯π₯ =
β8 Β± β4 Γ β1110 ππππππππππππππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β8 Β± 2β11
10
π₯π₯ = β 910
+ β4110
and π₯π₯ = β 910β β41
10 π₯π₯ = β 8
10+ 2β11
10 and π₯π₯ = β 8
10β 2β11
10
which reduces to
π₯π₯ = β45
+ β1110
and π₯π₯ = β45β β11
10
In the left hand column, we could have done the reducing without separating into two fractions:
π₯π₯ =β8 Β± β4 Γ β11
10 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
β8 Β± 2β1110
π€π€βππππβ ππππππππππππππ π‘π‘ππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
β4 Β± β1110
But it is important to realize that all three integers --- the β8, the 2, and the 10 --- must ALL have a common factor in order to do this reduction. If not, then itβs better to split into two fractions first, and then reduce to lowest terms
1. Copy each equation into your notes, then apply the Quadratic Formula to solve for π₯π₯: a. 3π₯π₯2 β 5π₯π₯ β 1 = 0 b. 3π₯π₯2 β 6π₯π₯ β 1 = 0
* part A should not simplify, but part B should
Example 2 Whether a quadratic equation will have an answer where the radical simplifies, or whether it will not, depends on how the values of ππ2 and 4ππππ combine. Because 4ππππ is already a multiple of 4, then if ππ2 is also a multiple of 4, then we are guaranteed to have something that is reducible. In example 1, we contrasted 5π₯π₯2 + 9π₯π₯ + 2 = 0 with 5π₯π₯2 + 8π₯π₯ + 2 = 0. In the second equation, because the middle term is even, ππ2 is a multiple of 4, and the calculations underneath the radical symbol ended up as another multiple of 4, and therefore could be reduced. Thatβs not the only way that you can end up with an answer involving a mixed radical, but it is the most likely way. But itβs also possible that the radicand simplifies to an integer, instead of to a mixed radical like 2β11. Hereβs an example:
5π₯π₯2 + 9π₯π₯ β 2 = 0
Page 58 Page 59 Page 59 Page 65 Page 70
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 14 ALGEBRA I
Lesson 14: Using the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
identify ππ, ππ, and ππ
5π₯π₯2 + 9π₯π₯ β 2 = 0
ππ = 5 ππ = 9 ππ = β2
calculate ππ2 and 4ππππ
ππ2 = 81
4ππππ ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½4(5)(β2)
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β40
write out the Quadratic Formula (this is optional) π₯π₯ =
βππ Β± βππ2 β 4ππππ2ππ
substitute into the formula π₯π₯ =
β9 Β± β81 ββ40 2(5)
combine the #βs under the square root symbol AND under the fraction bar π₯π₯ =
β9 Β± β121 10
This time around, the radicand ended up as a perfect square. The square root of 121 is just 11 --- not a radical, but an integer. So the problem now involves no square root symbols:
π₯π₯ =β9 Β± β121
10
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
β9 Β± 1110
This means that the answers will end up as either a rational number, like this:
π₯π₯ =β9 + 11
10
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
210
π€π€βππππβ ππππππππππππππ π‘π‘ππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
1 5
Or an integer, like this:
π₯π₯ =β9β 11
10
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β
2010
π€π€βππππβ ππππππππππππππ π‘π‘ππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β2
2. Copy the equation into your notes, then use the Quadratic Formula to solve the equation. a. 4π₯π₯2 + 4π₯π₯ β 3 = 0 b. 6π₯π₯2 β 7π₯π₯ β 10 = 0
Example 3 It can be helpful to predict the type of answers that a quadratic equation will have. Knowing that the answers will NOT involve radicals is particularly important, because it means that the equation is factorable. Solving a quadratic from its factors is usually faster than any other method. The equation in example 2 was factorable, like this:
Page 58 Page 71
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 14 ALGEBRA I
Lesson 14: Using the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
5π₯π₯2 + 9π₯π₯ β 2 = 0 πππππππ‘π‘ππππππ ππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ (5π₯π₯ β 1)(π₯π₯ + 2)
And from the factors we can see why the two answers are 15
and β 2:
5π₯π₯ β 1 = 0 π€π€βππππ π₯π₯ = 15
π₯π₯ + 2 = 0 π€π€βππππ π₯π₯ = β2 Being able to predict what type of answers a quadratic equation will have depends on finding the value that will be underneath the square root symbol. This radicand decides the style of answer because it determines whether the square root will simplify to a mixed radical or to an integer. The radicand is based on the value we get when calculating ππ2 β 4ππππ. We call this portion of the Quadratic Formula the discriminant because it discriminates (which in this case means to separate into categories) between several types of answers. if the discriminant isβ¦ then the answers areβ¦ a perfect square ---
like 25 or 49 or 121 two rational numbers ---
two fractions, two integers, or one of each
positive, but not a perfect square--- like 24 or 50 or 120
two mixtures of rational and irrational numbers ---
like 4 5
+ β3 5
or β5Β±β172
zero one rational number --- fraction or integer
negative not possible as real numbers --- there are no answers
Here is an example of each type for you to try:
3. Copy the equation, then calculate the discriminant. When done, list the number and type of solutions that the quadratic equation will have.
a. 10π₯π₯2 β 7π₯π₯ + 1 = 0
b. 9π₯π₯2 + 12π₯π₯ + 4 = 0
c. 6π₯π₯2 β 5π₯π₯ β 3 = 0
d. 2π₯π₯2 β π₯π₯ + 7 = 0 Partial answers: a) rational b) one answer only c) mixtures of rational/irrational d) zero answers
Page 72
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 14 ALGEBRA I
Lesson 14: Using the Quadratic Formula Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Problem Set Use the Quadratic Formula:
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ with πππ₯π₯2 + πππ₯π₯ + ππ = 0
to solve each equation.
1. 3π₯π₯2 + 8π₯π₯ β 1 = 0
2. 2π₯π₯2 β 10π₯π₯ + 7 = 0
3. 5π₯π₯2 β 2π₯π₯ β 3 = 0
4. π₯π₯2 + 4π₯π₯ β 2 = 0
5. 4π₯π₯2 + 6π₯π₯ + 2 = 0
6. 7π₯π₯2 β 8π₯π₯ + 1 = 0
Page 73
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 15: Graphing from Standard Form
Example 1 The last few lessons focused on solving a quadratic equation using different methods. But what are we finding when we do that? The difference between a function and an equation is important. A function is a way to describe all of the possible values generated by a particular set of arithmetic operations. An equation takes that function and assigns it to one specific value. For example, consider this linear function and linear equation:
ππ = ππππ β ππ ππππ β ππ = ππππ this is a function this is an equation
that describes a line that says the height of the line is 11 you use it to generate ordered pairs you use it to find the x-coordinate
like (2,β1) and (7, 9) of just that point: ( _____ , 11) The same sort of thing is true for quadratic functions and quadratic equations:
ππ(ππ) = ππππ + ππππ β ππ ππππ + ππππ β ππ = ππ this is a function this is an equation
it describes a quadratic it says the quadratic has a height of zero you use it to generate ordered pairs you use it to find the π₯π₯-intercepts
like (1,β5) and (β3,β5) which are (β4, 0) and (2, 0) How do we know that these points are the π₯π₯-intercepts? Take a look:
π₯π₯2 + 2π₯π₯ β 8 = 0 Factors as: (π₯π₯ + 4)(π₯π₯ β 2) = 0 Which means that: π₯π₯ + 4 = 0 or π₯π₯ β 2 = 0 And solves as: π₯π₯ = β4 or π₯π₯ = 2 The underlying function is: ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ β 8 Watch what happens if we plug in β4 for π₯π₯:
ππ(β4) = (β4)2 + 2(β4) β 8 = 16 β 8 β 8 = 0 So when π₯π₯ = β4, the functionβs height is zero. That only happens when itβs crossing the π₯π₯-axis.
By the end of this lesson, you should be able to: β’ graph a quadratic function by finding the vertex from the ππ-intercepts β’ graph a quadratic function by finding the vertex by completing the square
Page 74
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
The same is true when = 2 : ππ(2) = (2)2 + 2(2) β 8
= 4 + 4 β 8 = 0
1. Each quadratic equation is factored. Use the factors to find the π₯π₯-intercepts of the function: a. π₯π₯2 β 5π₯π₯ β 24 = 0
(π₯π₯ β 8)(π₯π₯ + 3) = 0 For ππ(π₯π₯) = π₯π₯2 β 54 β 24 The π₯π₯-intercepts are:
b. π₯π₯2 + 6π₯π₯ + 5 = 0 (π₯π₯ + 5)(π₯π₯ + 1) = 0 For ππ(π₯π₯) = π₯π₯2 = 6π₯π₯ + 5 The π₯π₯-intercepts are:
Example 2 So what can you do with the π₯π₯-intercepts? Well, you know from Module 3 that quadratic functions are symmetric --- the right hand half is a mirror image of the right hand half. And the line that goes through the middle of the two halves will also go through the vertex. When you graphed quadratic functions in Module 3, Topic C, you used the vertex form. For example, the function ππ(π₯π₯) = (π₯π₯ β 3)2 β 4 has its vertex at the point (3,β4) and opens up, like this: As you can see from the graph, its π₯π₯-intercepts are the points (1,0) and (5,0). And halfway between π₯π₯ = 1 and π₯π₯ = 5 is the value 3. The vertexβs π₯π₯-coordinate is the midpoint between the π₯π₯-intercepts:
1 + 52 ππππππππππππππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 6 2
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 3
What this all means is that one way to find the vertex of a quadratic equation is to find the π₯π₯-intercepts. Once we average the intercepts, we will know the first half of the vertexβs ordered pair. To find the second half, the π¦π¦-coordinate, we can then plug the π₯π₯-coordinate into the function.
Page 58 Page 59 Page 59 Page 75
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Hereβs what the whole process would look like: ππ(π₯π₯) = π₯π₯2 + 6π₯π₯ β 7
Set the function equal to zero: π₯π₯2 + 6π₯π₯ β 7 = 0 Factors as: (π₯π₯ + 7)(π₯π₯ β 1) = 0 Solves as: π₯π₯ = β7 ππππ π₯π₯ = 1 So the π₯π₯-intercepts are: (β7, 0) and (1,0) And the π₯π₯-coordinate of the vertex is: β7+1
2
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β 6
2
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ β3
Which means the π¦π¦-coordinate of the vertex is: (β3)2 + 6(β3) β 7 ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 9 β 18 β 7
πππππππππππποΏ½οΏ½οΏ½οΏ½οΏ½ β16
And now we plot the vertex of (β3,β16) as well as the π₯π₯-intercepts: To find more points, we can use the quadratic pattern off of the vertex, either as a table of values or as a series of movements, which we then connect into a smooth curve:
2. Copy the equation into your notes. Factor it, then identify the π₯π₯-intercepts. Use the intercepts to find the vertex.
a. π₯π₯2 + 4π₯π₯ β 5 = 0 b. π₯π₯2 β 6π₯π₯ + 8 = 0
Page 76
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 3 Not every quadratic function will be factorable. We have seen that in previous lessons in this module. So how else can we find the vertex and graph the shape? Another option besides factoring is to complete the square. You saw this process back in Lesson 11. Here is an example:
ππ(π₯π₯) = π₯π₯2 β 8π₯π₯ + 10 This is not factorable --- there are no factor pairs for 10 that would add to β8. But we can complete the square fairly easily, since the middle number is an even number:
ππ(π₯π₯) = π₯π₯2 β 8π₯π₯ + 10 β10 β 10
ππ(π₯π₯) β 10 = π₯π₯2 β 8π₯π₯ Γ· 2
ππ(π₯π₯) β 10 = (π₯π₯ β 4)2
+16 ππ(π₯π₯) + 6 = (π₯π₯ β 4)2
ππ(π₯π₯) = (π₯π₯ β 4)2 β 6 Now we know that this quadratic has a vertex of (4,β6) and opens up: Here is an example for you to try in your notes:
3. Complete the square, then graph the quadratic: ππ(π₯π₯) = π₯π₯2 + 6π₯π₯ + 11 = 0
Did you get a vertex of (β3,β2)? Congratulations!
Page 77
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Example 4 Of course, not every quadratic lends itself to an easy complete-the-square process, either. For example, the middle term might not divide easily by 2, or there might be a coefficient other than 1 on the leading term. In those situations, we turn instead to using part of the quadratic formula. The Quadratic Formula looks like this:
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ with πππ₯π₯2 + πππ₯π₯ + ππ = 0
and the first portion of it is also the formula for finding the π₯π₯-coordinate of the vertex of the quadratic. We usually refer to the vertex as (β,ππ), so the formula for finding the π₯π₯-coordinate is:
β = β ππ2ππ
And then, to find the ππ value, we plug the β value back into the function. Hereβs an example of how the process would look:
ππ(π₯π₯) = 2π₯π₯2 β 8π₯π₯ + 7 ππ = 2, ππ = β8, ππ = 7 β = β ππ
2ππ ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ +8
2(2) ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 8
4
ππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 2
ππ = ππ(2) π€π€βππππβ ππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 2(2)2 β 8(2) + 7
ππππππππππππππ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ 2(4) β 16 + 7
πππππππππππποΏ½οΏ½οΏ½οΏ½οΏ½ β1
We now know the vertex is the point (2,β1) and that the quadratic has been stretched by a factor of 2: Here are two practice problems for you to try in your notes:
4. Use the formula β = β ππ2ππ
to find the π₯π₯-coordinate of the vertex. Then plug that value into the function to find the π¦π¦-coordinate. You do not need to graph the function.
a. ππ(π₯π₯) = π₯π₯2 + 5π₯π₯ β 4 = 0 (the vertex will have fraction coordinates)
b. ππ(π₯π₯) = βπ₯π₯2 β 6π₯π₯ + 1
Page 78
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Problem Set Find the vertex of each quadratic function: 1. ππ(π₯π₯) = 3π₯π₯2 + 6π₯π₯ β 1
2. ππ(π₯π₯) = βπ₯π₯2 β 10π₯π₯ β 17
3. ππ(π₯π₯) = π₯π₯2 β 2π₯π₯ β 3
4. ππ(π₯π₯) = π₯π₯2 + 4π₯π₯ β 2
5. ππ(π₯π₯) = βπ₯π₯2 + 6π₯π₯ + 2
6. ππ(π₯π₯) = π₯π₯2 β 8π₯π₯ + 1
Find the vertex and graph the function: 7. ππ(π₯π₯) = π₯π₯2 β 4π₯π₯ + 3
Page 76 Page 79
NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 15 ALGEBRA I
Lesson 15: Graphing from Standard Form Last edited on 3/5/18
Β© 2013 Common Core, Inc. Some rights reserved. commoncore.org
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
8. ππ(π₯π₯) = βπ₯π₯2 + 6π₯π₯ β 2
9. ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ β 6
10. ππ(π₯π₯) = 2π₯π₯2 + 12π₯π₯ + 9
Page 80