coming up in math 110: today: section 8.2 (quadratic formula) tomorrow: review for quiz 4. note:...

Coming up in Math 110:

• Today: Section 8.2 (Quadratic formula)• Tomorrow: Review for Quiz 4.

NOTE: Practice Quiz 4 is open now and ready for you to take as many times as you want.

• Wednesday: Quiz 4 on sections 5.5 - 5.8 & 8.2 (Factoring polynomials and solving polynomial equations.)

• Thursday: Review for Test 2

Next week:

Monday: Test 2 (100-point test on all of Chapter 5 + section 8.2)

NOTE: Because this test occurs on a Monday, there will be an optional review session on Sunday, March 30th. That session will be held from 7:00-9:00 p.m. in this classroom (214) and it is open to all Math 110 students from all sections.

Any questions on the Section 5.8 homework?

Pass your worksheets for this assignment to the middle

aisle for pickup now.

Remember the problem like this one from the homework that was due today?

Wouldn’t it be nice if there was an easier way to do it than by factoring?

Leave factoring up on board: (4x - 9)(3x + 8)

Please

CLOSE

YOUR LAPTOPS,and turn off and put away your

cell phones,

and get out your note-taking materials.

Section 8.2

The Quadratic Formula

The Quadratic Formula

The quadratic formula is another technique we can use for solving quadratic equations.

Remember, quadratic equations are polynomial equations of degree 2, such as

x2 + 3x -7 = 0

or

5x2 – 14 = 0.

The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. – See Section 8.1 if you’re interested in seeing how

this formula is derived. – This will also be covered in Math 120 in more

detail, along with the technique called “completing the square”.

The Quadratic Formula:

a

acbbx

2

42

The solutions to the equation ax2 + bx + c = 0

are given by the formula

Note: This formula IS on the pink formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.

The Big Question:

How can we tell when we should use

factoring and when we should use the quadratic formula?

Solve x2 + 4x + 3 = 0 by

• Factoring• The quadratic formula.

Which way works best?

Example 1

Solve x2 + 4x + 3 = 0 by Factoring:

This one is pretty easy to factor. The factoring is (x + 3)(x + 1) = 0, so the solutions are given by x + 3 = 0, or x = -3,

and x + 1 = 0, which gives x = -1.

Now, solve x2 + 4x + 3 = 0 by the quadratic formula:

a = 1, b = 4, c = 3, so the formula gives:

32

6

2

24

12

2

2

24

2

24

2

44

2

12164

12

31444 2

or

x

Which way works best for this problem?

In this case, the factoring method is much quicker, although BOTH methods give the same answer.

Solve x2 + 5x + 12 = 0 by

• Factoring• The quadratic formula.

Which way works best?

Example 2

Solve x2 + 5x + 12 = 0 by Factoring: This one looks pretty easy to factor, but when you

start trying to find two factors of 12 that add up to 5, nothing works.

(1+12=13, 2+6=8, 3+4=7).

What does this mean? It means that the polynomial is PRIME, and there

are no rational solutions. (Remember, a rational number is either an integer or a fraction.)

Solve x2 + 5x + 12 = 0 (continued):

• Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12

so the formula gives:

2

235

2

48255

12

121455 2

x

Notice that the number under the radical sign is negative, which means there are no real answers. If the number under the square root sign comes out to be positive but it’s not a perfect square, this means the answer is a real number, but is irrational because it can’t be simplified to remove the radical. In either of these cases, we’d say the polynomial is prime, and therefore has no rational roots.

So which way works best for solving

x2 + 5x + 12 = 0?

Either way works fine, but if you think a polynomial is prime, a good way to check is by calculating the discriminant (b2 – 4ac). If the discriminant is either negative or not a perfect square, then you know for sure that your polynomial is prime and there are no rational solutions.

Solve x2 + 2x - 4 = 0 by

• Factoring• The quadratic formula.

Which way works best?

Example 3

Solve x2 + 2x - 4 = 0 by Factoring: Again, this one looks like it would be easy to factor,

but when you start trying to find two factors of 4 that add up to 2, nothing works.

(-1 + 4 = 3, -4 + 1=-3, -2+2 = 0)

What does this mean? It means that the polynomial is PRIME, and there

are no rational solutions. (Remember, a rational number is either an integer or a fraction.)

Solve x2 + 2x - 4 = 0 (continued):

• Let’s see what the quadratic formula gives in this case: a = 1, b = 2, c = -4

so the formula gives:

2

202

2

1642

12

)4(1422 2

x

Notice that the number (20) under the radical sign is positive, which means there are two real answers. However, since 20 is not a perfect square, its square root is irrational. In this case, we’d again say that the polynomial is prime, because it has no rational roots.

Solve x2 + 2x - 4 = 0 using the quadratic formula (continued):

2

202

2

1642

12

)4(1422 2

x

IMPORTANT NOTE: Because the discriminant (20) is POSITIVE, we still do have two REAL NUMBER solutions, but they won’t be rational numbers. How do we find these two real answers if we can’t factor the polynomial?

20

So, if you were asked to find the approximate solution(s) of this equation rounded to the nearest hundredth, you’d use your calculator to estimate . Do this now and see what you get for your two answers. Answers: x = 1.24 and x = -3.24. (How would you check this? Will you get EXACTLY zero?)

20

Answer: continue calculating the answer using the quadratic formula, using your calculator to give you a decimal approximation of .

Now re-do this problem from the 5.8 homework using the quadratic formula:

Answers: -8/3, 9/4

Do this on your calculator now, showing work in your notebook.• What is the discriminant?• Is it positive? • Is it a perfect square? • How many real answers will there be? • Will they be rational or irrational?

Recap: For the equation 12x2 + 5x -72 = 0, should we solve it using the factoring approach or the quadratic formula?

Either way works, but the quadratic formula approach is probably going to be faster than factoring for most people.

Moral of the story: For a quadratic equation with a leading coefficient other than 1, it’s probably going to be quicker to solve it using the quadratic formula than it would be to factor the polynomial.

)1(2

)20)(1(4)8(8 2

x

2

80648

2

1448

2

128210,

2

4

2

20 or or

x2 + 8x – 20 = 0 (multiply both sides by 8)

a = 1, b = 8, c = -20

8

1

2

5Solve x2 + x – = 0 by the quadratic formula.

Question: What if some coefficients in your quadratic equation are fractions? ANSWER: Clear them first by multiplying all terms by the LCD:

• The expression under the radical sign in the quadratic formula (b2 – 4ac) is called the discriminant.

• The discriminant will take on a value that is positive, 0, or negative.

• The value of the discriminant indicates two distinct real solutions (if it’s positive), one real solution (if it’s zero), or two complex, but not real solutions (if it’s negative – a topic to be discussed in Math 120).

No x-intercepts

No real solution; two complex imaginary solutions

b2 – 4ac < 0

One x-intercept

One real solution (a repeated solution)(If b2 – 4ac is a perfect square, the solution will be a rational number. If not, it’s irrational.)

b2 – 4ac = 0

Two x-intercepts

Two unequal real solutions(If b2 – 4ac is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.)

b2 – 4ac > 0

Graph of y = ax2 + bx + c

Kinds of solutions to ax2 + bx + c = 0

Discriminantb2 – 4ac

The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0

Use the discriminant to determine the number and type of solutions for the following equation.

5 – 4x + 12x2 = 0

a = 12, b = -4, and c = 5

b2 – 4ac = (-4)2 – 4(12)(5)

= 16 – 240

= -224

Since the discriminant is negative, there are no real solutions.

Question: What would this graph look like?

Example

Use the discriminant to determine the number and type of solutions for the following equation.

25x2 - 4 = 0

a = 25, b = 0 (why?) , and c = -4

b2 – 4ac = (0)2 – 4(25)(-4) = 0 – -400 = 400

Since the discriminant is positive, there are two real solutions.

(You could go on to show that the solutions are 2/5 and -2/5, either by factoring or using the quadratic formula.)

Example

Use the discriminant to determine the number and type of solutions for the following equation.

5 – 4x + 12x2 = 0

a = 12, b = -4, and c = 5

b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 24 = -224

Since the discriminant is negative, there are no real solutions.

NOTE: There will be two COMPLEX solutions that are not real numbers, i.e. they will contain the imaginary number i which is the square root of -1.

Example

Use the discriminant to determine the number and type of solutions for the following equation.

x2 – 8x + 16 = 0

a = 1, b = -8, and c = 16

b2 – 4ac = (-8)2 – 4(1)(16) = 64 – 64 = 0

Since the discriminant is zero, there is one real solution. (You could go on to show that the solution is 4, either by factoring or using the quadratic formula.)

Question: What would this graph look like?

Example

2 possible approaches:

1. Exact answer: The exact answer will contain a radical, i.e. it will be an irrational number. (More on this in Chapter 7...)

2. Approximate answer: Use your calculator to get an approximate decimal answer.

How do you figure out the answers if the discriminant is positive but not a perfect square?

REMINDER!!!IMPORTANT NOTE:

Use the quadratic formula technique to solve all problems in this homework assignment.

There are a couple of word problems at the end of the assignment in which the online learning aids will show factoring as the solution method. You should use the quadratic formula instead (and you will find it to be easier and quicker than factoring.)

Reminder:

This homework assignment on Section 8.2 is due

at the start of next class period.

You may now OPEN

your LAPTOPSand begin working on the

homework assignment (if there’s any time left...)

But remember, you can always work in the JHSW 203 open lab after class (or before

tour next class session) if you want some help on this homework.