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Page 1: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Section 5: Quadratic Equations and Functions – Part 1

Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Let’s revisit linear functions. Imagine that you are driving down the road at a constant speed of 40mph. This is a linear function.

We can represent the distance traveled versus time on a table (to the right).

Time (in hours)

Distance Traveled (in miles)

1 402 803 1204 160

We can represent the scenario on a graph.

Page 2: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

We can represent the distance traveled 𝑑(𝑑), in terms of time, 𝑑 hours, with the equation 𝑑(𝑑) = 40𝑑. Linear functions always have a constant rate of change. In this section, we are going to discover a type of non-linear function. Consider the following situation. Liam dropped a watermelon from the top of a 300 ft tall building. He wanted to know if the watermelon was falling at a constant rate over time. He filmed the watermelon’s fall and then recorded his observations in the following table.

Time (in seconds)

Height (in feet)

0 300.01 283.92 235.63 155.14 42.4

What do you notice about the rate of change? It is not constant. Why do you think that the rate of change is not constant? Gravity causes the watermelon to get faster as it gets closer to the ground.

Page 3: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Liam entered the data of the falling watermelon into his graphing calculator. The graph below displays the first quadrant of the graph.

What is the independent variable? Time What is the dependent variable? Height Liam then used his calculator to find the equation of the function, β„Ž 𝑑 = βˆ’16𝑑3 + 300.

Page 4: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Important facts:

Ø We call this non-linear function a ___________________. Ø The general form of the equation is

_________________________. The graph of 𝑓(π‘₯) = π‘₯3 is shown below.

Ø This graph is called a ________________________. Why did we only consider the first quadrant of Liam’s graph? You wouldn’t have negative time or height

quadratic

𝒇(𝒙) = π’‚π’™πŸ + 𝒃𝒙 + 𝒄

parabola

Page 5: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

In Liam’s graph, what was the watermelon’s height when it hit the ground? 𝟎 feet The time when the watermelon’s height was at zero is called the solution to this quadratic equation. We also call this the ____________ of the equation. There was only one solution to Liam’s equation. Describe a situation where there could be two solutions. Answers vary. Sample answer: A rocket’s height over time after is launched from the ground and lands back on the ground. What about no solutions? Answers vary. Sample answer: A bungee jumpers height over time as they jump from a 𝟐𝟎𝟎foot platform.

To solve a quadratic equation using a graph:

Ø Look for the ________________________ of the graph.

Ø The solutions are the values where the graph intercepts the ________________________.

zeros

𝒙-intercepts

𝒙-axis

Page 6: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Let’s Practice! 1. What are the solutions to the quadratic equation graphed

below?

𝒙 = 𝟐 and 𝒙 = πŸ“

Ø

Zeros = π‘₯-intercepts = Solutions

Page 7: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Try It! 2. Aaron shot a water bottle rocket from the ground. A

graph of height over time is shown below.

a. What type of function best models the rocket’s motion? quadratic

b. After how many seconds did the rocket hit the

ground? πŸ“

c. Estimate the maximum height of the rocket.

about πŸ”. πŸπŸ“ meters The maximum or minimum point of the parabola is called the ________________. vertex

Page 8: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

BEAT THE TEST!

1. Jordan owns an electronics business. During her first year in the business, she collected data and created the following graph showing the relationship between the selling price of an item and the profit.

Part A: Circle the solutions to the quadratic function

graphed above. Part B: What do the solutions represent?

The prices that yielded NO profit.

Part C: Box the vertex of the graph. Part D: What does the vertex represent?

The price that yields the maximum profit.

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Page 9: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Section 5 – Topic 2 Factoring Quadratic Expressions

Let’s review the two methods we used for multiplying polynomials. Area Model:

Distributive Property: 3 π‘₯ + 2𝑦 βˆ’ 7𝑧 = πŸ‘π’™ + πŸ”π’š βˆ’ πŸπŸπ’›

We can use these same methods to factor out the greatest common factor of an expression. Area Model:

Distributive Property: 10π‘₯G βˆ’ 14π‘₯3 + 12π‘₯ = πŸπ’™(πŸ“π’™πŸ βˆ’ πŸ•π’™ + πŸ”)

!"" #$ βˆ’&'(

) 2+ βˆ’7-

3

10#$% βˆ’14#( 12#

*%+ βˆ’,% -

+%

Page 10: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Use the area model to write an equivalent expression for (2π‘₯ + 5)(π‘₯ + 3).

πŸπ’™πŸ + πŸπŸπ’™ + πŸπŸ“ We can use this same area model to factor a quadratic expression. Look at the resulting trinomial and notice the following four patterns:

Ø The first term of the trinomial can always be found in the ________________ ______________ rectangle.

Ø The last term of the trinomial can always be found in

the ________________ ______________ rectangle. Ø The second term of the trinomial is the ____________ of

the ________________ _____________ and ____________ ______________ rectangles.

Ø The ________________ of the __________________ are

always equal.

πŸπ’™πŸ β‹… πŸπŸ“ = πŸ‘πŸŽπ’™πŸ and πŸ“π’™ β‹… πŸ”π’™ = πŸ‘πŸŽπ’™πŸ

2"

5

" 3

%&% '&

(& )(

top left

bottom right

bottom left top

right

sum

product diagonals

Page 11: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Use the distributive property to write an equivalent expression for (2π‘₯ + 5)(π‘₯ + 3).

πŸπ’™ + πŸ“ 𝒙 + πŸ‘ = πŸπ’™πŸ + πŸ”π’™ + πŸ“π’™ + πŸπŸ“ = πŸπ’™πŸ + πŸπŸπ’™ + πŸπŸ“ We can also use the distributive property to factor a quadratic expression. What are the two middle terms of the expanded form? πŸ”π’™and πŸ“π’™

Consider the resulting trinomial. 2π‘₯3 + 11π‘₯ + 15 Notice that the product of the two middle terms of expanded form are equal to the product of the first and last term of the trinomial. The middle terms also sum to the middle term of the trinomial. πŸ”π’™ βˆ™ πŸ“π’™ = πŸ‘πŸŽπ’™πŸ and πŸ”π’™ + πŸ“π’™ = πŸπŸπ’™ Let’s consider how we can use this and the distributive property to factor a quadratic expression.

Page 12: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Factor 2π‘₯3 + 3π‘₯ βˆ’ 5 using the distributive property. Ø Multiply the first term by the last term.

πŸπ’™πŸ βˆ™ βˆ’πŸ“ = βˆ’πŸπŸŽπ’™πŸ

Ø Find two factors whose product is equal to βˆ’10π‘₯3 and whose sum is equal to 3π‘₯.

πŸ“π’™ and – πŸπ’™

Ø Replace the middle term with these two factors. πŸπ’™πŸ + πŸ“π’™ βˆ’ πŸπ’™ βˆ’ πŸ“

Ø Factor the polynomial by grouping the first 2 terms and the last 2 terms.

πŸπ’™πŸ + πŸ“π’™ + (βˆ’πŸπ’™ βˆ’ πŸ“) 𝒙 πŸπ’™ + πŸ“ βˆ’ 𝟏(πŸπ’™ + πŸ“)

(πŸπ’™ + πŸ“)(𝒙 βˆ’ 𝟏)

Page 13: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Let’s Practice! 1. Consider the quadratic expression 3π‘₯3 + 4π‘₯ βˆ’ 4.

a. Factor using the area model.

(𝒙 + 𝟐)(πŸ‘π’™ βˆ’ 𝟐)

b. Factor using the distributive property.

πŸ‘π’™πŸ + πŸ’π’™ βˆ’ 𝟐 = πŸ‘π’™πŸ βˆ’ πŸπ’™ + πŸ”π’™ βˆ’ πŸ’

= πŸ‘π’™πŸ βˆ’ πŸπ’™ + πŸ”π’™ βˆ’ πŸ’ = 𝒙 πŸ‘π’™ βˆ’ 𝟐 + 𝟐 πŸ‘π’™ βˆ’ 𝟐 = (πŸ‘π’™ βˆ’ 𝟐)(𝒙 + 𝟐)

!

"

#! βˆ’"

#!" βˆ’"!

%! βˆ’&

Ø

You can check your answer by using the distributive property. The product of the factors should always result in the original trinomial.

Page 14: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Try It! 2. Consider the quadratic expression 4𝑀3 βˆ’ 21𝑀 + 20.

a. Factor using the area model.

(πŸ’π’˜ βˆ’ πŸ“)(π’˜ βˆ’ πŸ’)

b. Factor using the distributive property. πŸ’π’˜πŸ βˆ’ πŸπŸπ’˜ + 𝟐𝟎

= πŸ’π’˜πŸ βˆ’ πŸ“π’˜ βˆ’ πŸπŸ”π’˜ + 𝟐𝟎 = πŸ’π’˜πŸ βˆ’ πŸ“π’˜ + βˆ’πŸπŸ”π’˜ + 𝟐𝟎

= π’˜ πŸ’π’˜βˆ’ πŸ“ βˆ’ πŸ’ πŸ’π’˜ βˆ’ πŸ“ = (πŸ’π’˜ βˆ’ πŸ“)(π’˜ βˆ’ πŸ’)

!"

βˆ’$

" βˆ’!

!"% βˆ’&'"

βˆ’$" %(

Page 15: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

BEAT THE TEST!

1. Identify all factors of the expression 18π‘₯3 βˆ’ 9π‘₯ βˆ’ 5. Select all that apply.

Β¨ 2π‘₯ + 5 Γ½ 6π‘₯ βˆ’ 5 Β¨ 18π‘₯ βˆ’ 5 Β¨ 3π‘₯ + 5 Γ½ 3π‘₯ + 1

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Page 16: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Section 5 – Topic 3 Solving Quadratic Equations by Factoring

Once a quadratic equation is factored, we can use the zero product property to solve the equation.

The zero product property states that if the product of two factors is zero, then one (or both) of the factors must be _________________.

Ø If π‘Žπ‘ = 0, then either π‘Ž = 0, 𝑏 = 0,or π‘Ž = 𝑏 = 0.

To solve a quadratic equation by factoring:

Step 1: Set the equation equal to zero. Step 2: Factor the quadratic. Step 3: Set each factor equal to zero and solve. Step 4: Write the solution set.

zero

Page 17: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Let’s Practice! 1. Solve for 𝑏 by factoring 𝑏3 + 8𝑏 + 15 = 0.

π’ƒπŸ + πŸ“π’ƒ + πŸ‘π’ƒ + πŸπŸ“ = 𝟎 𝒃 𝒃 + πŸ“ + πŸ‘ 𝒃 + πŸ“ = 𝟎 𝒃 + πŸ“ 𝒃 + πŸ‘ = 𝟎 𝒃 + πŸ“ = 𝟎 or 𝒃 + πŸ‘ = 𝟎 𝒃 = βˆ’πŸ“ or 𝒃 = βˆ’πŸ‘

{βˆ’πŸ“,βˆ’πŸ‘} 2. Solve for 𝑓 by factoring 10𝑓3 + 17𝑓 + 3 = 0.

πŸ“π’‡ + 𝟏 πŸπ’‡ + πŸ‘ = 𝟎 πŸ“π’‡ + 𝟏 = 𝟎 or πŸπ’‡ + πŸ‘ = 𝟎 𝒇 = βˆ’πŸ

πŸ“ or 𝒇 = βˆ’πŸ‘

𝟐

βˆ’πŸ‘πŸ ,βˆ’

πŸπŸ“

!"

#

$" %

#&"$ #!"

$" %

Page 18: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Try It! 3. Solve for 𝑗 by factoring 6𝑗3 βˆ’ 19𝑗 + 14 = 0.

πŸ”π’‹ βˆ’ πŸ• 𝒋 βˆ’ 𝟐 = 𝟎 πŸ”π’‹ βˆ’ πŸ• = 𝟎 or 𝒋 βˆ’ 𝟐 = 𝟎 𝒋 = πŸ•

πŸ” or 𝒋 = 𝟐

πŸ•πŸ” , 𝟐

!"

βˆ’$

" βˆ’%

!"% βˆ’&%"

βˆ’$" &'

Page 19: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

BEAT THE TEST!

1. Tyra solved the quadratic equation π‘₯3 βˆ’ 10π‘₯ βˆ’ 24 = 0 by factoring. Her work is shown below:

Step 1: π‘₯3 βˆ’ 10π‘₯ βˆ’ 24 = 0 Step 2: π‘₯3 βˆ’ 4π‘₯ βˆ’ 6π‘₯ βˆ’ 24 = 0 Step 3: π‘₯3 βˆ’ 4π‘₯ + (βˆ’6π‘₯ βˆ’ 24) = 0 Step 4: π‘₯ π‘₯ βˆ’ 4 βˆ’ 6(π‘₯ βˆ’ 4) = 0Step 5: (π‘₯ βˆ’ 4)(π‘₯ βˆ’ 6) = 0 Step 6: π‘₯ βˆ’ 4 = 0, π‘₯ βˆ’ 6 = 0 Step 7: π‘₯ = 4 or π‘₯ = 6 Step 8: 4, 6

Tyra did not find the correct solutions. Investigate the steps, decipher her mistakes, and explain how to correct Tyra’s work.

Step 2: She replaced βˆ’πŸπŸŽπ’™ with βˆ’πŸ’π’™ and – πŸ”π’™. While their sum is – πŸπŸŽπ’™, their product is not – πŸπŸ’π’™πŸ. Step 4: When she factored out – πŸ”, she should have gotten – πŸ” 𝒙 + πŸ’ . π’™πŸ βˆ’ πŸπŸπ’™ + πŸπ’™ βˆ’ πŸπŸ’ = 𝟎 𝒙 𝒙 βˆ’ 𝟏𝟐 + 𝟐 𝒙 βˆ’ 𝟏𝟐 = 𝟎 𝒙 βˆ’ 𝟏𝟐 𝒙 + 𝟐 = 𝟎 𝒙 βˆ’ 𝟏𝟐 = 𝟎 and 𝒙 + 𝟐 = 𝟎 𝒙 = 𝟏𝟐 and 𝒙 = βˆ’πŸ

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Page 20: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Section 5 – Topic 4 Solving Other Quadratic Equations by Factoring

Many quadratic equations will not be in standard form. Ø The equation won’t always equal zero. Ø There may be a greatest common factor (GCF) within

all of the terms. Let’s Practice! 1. Solve for π‘š: 3π‘š3 + 30π‘š βˆ’ 168 = 0.

πŸ‘ π’ŽπŸ + πŸπŸŽπ’Žβˆ’ πŸ“πŸ” = 𝟎 πŸ‘ π’Ž+ πŸπŸ’ π’Žβˆ’ πŸ’

πŸ‘ =πŸŽπŸ‘

π’Ž+ πŸπŸ’ = 𝟎 or π’Žβˆ’ πŸ’ = 𝟎 π’Ž = βˆ’πŸπŸ’ or π’Ž = πŸ’

2. Solve for π‘₯: π‘₯ + 4 π‘₯ βˆ’ 5 = βˆ’8.

π’™πŸ βˆ’ πŸ“π’™ + πŸ’π’™ βˆ’ 𝟐𝟎 = βˆ’πŸ– π’™πŸ βˆ’ 𝒙 βˆ’ 𝟐𝟎 + πŸ– = 𝟎 π’™πŸ βˆ’ 𝒙 βˆ’ 𝟏𝟐 = 𝟎 𝒙 βˆ’ πŸ’ 𝒙 + πŸ‘ = 𝟎 𝒙 = πŸ’ or 𝒙 = βˆ’πŸ‘

Page 21: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Try It! 3. Solve for 𝑑: 6𝑑3 + 5𝑑 = 1. πŸ”π’…πŸ + πŸ“π’… βˆ’ 𝟏 = 𝟎

πŸ”π’… βˆ’ 𝟏 𝒅 + 𝟏 = 𝟎 𝒅 = 𝟏

πŸ” or 𝒅 = βˆ’πŸ

4. Solve for 𝑝: 𝑝3 + 36 = 13𝑝.

π’‘πŸ βˆ’ πŸπŸ‘π’‘ + πŸ‘πŸ” = 𝟎 𝒑 βˆ’ πŸ— 𝒑 βˆ’ πŸ’ = 𝟎 𝒑 βˆ’ πŸ— = 𝟎 or 𝒑 βˆ’ πŸ’ = 𝟎

𝒑 = πŸ— or 𝒑 = πŸ’

!"

βˆ’$

" $

!"% !"

βˆ’$" $

Page 22: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

BEAT THE TEST!

1. What are the solutions to 40π‘₯3 βˆ’ 30π‘₯ = 135? Select all that apply.

Β¨ βˆ’ ]

3 Β¨

G^

Β¨ βˆ’ ]^ Β¨

G3

Γ½ βˆ’32 Γ½ ]

^

Β¨ βˆ’G^

πŸ’πŸŽπ’™πŸ βˆ’ πŸ‘πŸŽπ’™ βˆ’ πŸπŸ‘πŸ“ = 𝟎 πŸ“ πŸ–π’™πŸ βˆ’ πŸ”π’™ βˆ’ πŸπŸ•

πŸ“ =πŸŽπŸ“

πŸ–π’™πŸ βˆ’ πŸ”π’™ βˆ’ πŸπŸ• = 𝟎

πŸπ’™ + πŸ‘ πŸ’π’™ βˆ’ πŸ— = 𝟎 πŸπ’™ + πŸ‘ = 𝟎 or πŸ’π’™ βˆ’ πŸ— = 𝟎 𝒙 = βˆ’πŸ‘

𝟐 or 𝒙 = πŸ—

πŸ’

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2!

"

#! βˆ’%

&!' βˆ’(&!

('! βˆ’')

Page 23: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Section 5 – Topic 5 Solving Quadratic Equations by Factoring –

Special Cases

There are a few special cases when solving quadratic equations by factoring. Perfect Square Trinomials:

Ø π‘₯3 + 6π‘₯ + 9 is an example of perfect square trinomial. We see this when we factor.

Ø A perfect square

trinomial is created when you square a ________________________.

Recognizing a Perfect Square Trinomial: A quadratic expression can be factored as a perfect square trinomial if it can be re-written in the form π‘Ž3 + 2π‘Žπ‘ + 𝑏3. Factoring a Perfect Square Trinomial:

Ø If π‘Ž3 + 2π‘Žπ‘ + 𝑏3 is a perfect square trinomial, then π‘Ž3 + 2π‘Žπ‘ + 𝑏3 = π‘Ž + 𝑏 3.

Ø If π‘Ž3 βˆ’ 2π‘Žπ‘ + 𝑏3 is a perfect square trinomial, then

π‘Ž3 βˆ’ 2π‘Žπ‘ + 𝑏3 = π‘Ž βˆ’ 𝑏 3.

!

"

! "

!# "!

"! $

binomial

Page 24: Section 5: Quadratic Equations and Functions – Part 1 · PDF fileSection 5: Quadratic Equations and Functions – Part 1 Section 5 – Topic 1 Real-World Examples of Quadratic Functions

Let’s Practice! 1. Determine whether 16π‘₯3 + 88π‘₯ + 121 is a perfect square

trinomial. Justify your answer.

It is a perfect square trinomial. It can be written in the form π’‚πŸ + πŸπ’‚π’ƒ + π’ƒπŸ.

πŸ’π’™ 𝟐 + 𝟐 πŸ’π’™ 𝟏𝟏 + 𝟏𝟏 𝟐 2. Solve for π‘ž: π‘ž3 βˆ’ 10π‘ž + 25 = 0.

𝒒 𝟐 βˆ’ 𝟐 𝒒 πŸ“ + πŸ“ 𝟐 = 𝟎 𝒒 βˆ’ πŸ“ 𝟐 = 𝟎 (𝒒 βˆ’ πŸ“) = 𝟎 or (𝒒 βˆ’ πŸ“) = 𝟎 𝒒 = πŸ“

πŸ“ Try It! 3. Determine whether π‘₯3 βˆ’ 8π‘₯ + 64 is a perfect square

trinomial. Justify your answer. It is NOT a perfect square trinomial. It cannot be written in the form π’‚πŸ + πŸπ’‚π’ƒ + π’ƒπŸ.

π’™πŸ βˆ’ πŸ–π’™ + πŸ”πŸ’ β‰  (𝒙)𝟐 βˆ’ 𝟐 𝒙 πŸ– + πŸ– 𝟐

4. Solve for 𝑀: 4𝑀3 + 49 = βˆ’28𝑀. πŸ’π’˜πŸ + πŸπŸ–π’˜ + πŸ’πŸ— = 𝟎 πŸπ’˜ 𝟐 + 𝟐(πŸπ’˜)(πŸ•) + πŸ• 𝟐 = 𝟎

πŸπ’˜ + πŸ• 𝟐 = 𝟎 π’˜ = βˆ’πŸ•

𝟐 βˆ’πŸ•

𝟐

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5. What do you notice about the number of solutions to perfect square quadratic equations? There is always one solution.

6. Sketch the graph of a quadratic equation that is a perfect

square trinomial. Sample Answer:

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Difference of Squares Use the distributive property to multiply the following binomials. (π‘₯ + 5)(π‘₯ βˆ’ 5) π’™πŸ βˆ’ πŸ“π’™ + πŸ“π’™ βˆ’ πŸπŸ“ π’™πŸ βˆ’ πŸπŸ“ (5π‘₯ + 3)(5π‘₯ βˆ’ 3) πŸπŸ“π’™πŸ βˆ’ πŸπŸ“π’™ + πŸπŸ“π’™ βˆ’ πŸ— πŸπŸ“π’™πŸ βˆ’ πŸ— Describe any patterns you notice. Answers vary. Sample answer: The middle terms always canceled out.

Ø When we have a binomial in the form π‘Ž3 βˆ’ 𝑏3, it is called the difference of two squares. We can factor this as π‘Ž + 𝑏 π‘Ž βˆ’ 𝑏 .

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Let’s Practice! 7. Solve the equation 49π‘˜3 = 64 by factoring. πŸ’πŸ—π’ŒπŸ βˆ’ πŸ”πŸ’ = 𝟎 πŸ•π’Œ βˆ’ πŸ– πŸ•π’Œ + πŸ– = 𝟎

πŸ•π’Œ βˆ’ πŸ– = 𝟎 or πŸ•π’Œ + πŸ– = 𝟎 π’Œ = πŸ–

πŸ• or π’Œ = βˆ’πŸ–πŸ•

βˆ’πŸ–πŸ•, πŸ–πŸ•

Try It! 8. Solve the equation 0 = 121𝑝3 βˆ’ 100.

πŸπŸπŸπ’‘πŸ βˆ’ 𝟏𝟎𝟎 = 𝟎 (πŸπŸπ’‘ βˆ’ 𝟏𝟎)(πŸπŸπ’‘ + 𝟏𝟎) = 𝟎

πŸπŸπ’‘ βˆ’ 𝟏𝟎 = 𝟎 or πŸπŸπ’‘ + 𝟏𝟎 = 𝟎 𝒑 = 𝟏𝟎

𝟏𝟏 or 𝒑 = βˆ’πŸπŸŽπŸπŸ

βˆ’πŸπŸŽπŸπŸ, 𝟏𝟎𝟏𝟏

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BEAT THE TEST!

1. Which of the following expressions are equivalent to 8π‘ŽG βˆ’ 98π‘Ž? Select all that apply. Γ½ 2(4π‘ŽG βˆ’ 49π‘Ž) Γ½ 2π‘Ž(4π‘Ž3 βˆ’ 49) Β¨ 2π‘Ž(4π‘ŽG βˆ’ 49π‘Ž) Β¨ 2π‘Ž βˆ’ 7 (2π‘Ž + 7) Β¨ 2 2π‘Ž βˆ’ 7 (2π‘Ž + 7) Γ½ 2π‘Ž 2π‘Ž βˆ’ 7 (2π‘Ž + 7)

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Section 5 – Topic 6 Solving Quadratic Equations by Taking Square Roots

Consider the following quadratic equation.

2π‘₯3 βˆ’ 36 = 0

When quadratic equations are in the form π‘Žπ‘₯3 + 𝑐 = 0, solve by taking the square root.

Step 1: Get the variable on the left and the constant on the right.

Step 2: Take the square root of both sides of the equation. (Don’t forget the negative root!)

Solve for π‘₯ by taking the square root. 2π‘₯3 βˆ’ 36 = 0 πŸπ’™πŸ

𝟐 =πŸ‘πŸ”πŸ

π’™πŸ = πŸπŸ– 𝒙 = Β± πŸπŸ– 𝒙 = Β±πŸ‘ 𝟐 or 𝒙 β‰ˆ Β±πŸ’. πŸπŸ’

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Let’s Practice! 1. Solve π‘₯3 βˆ’ 121 = 0. π’™πŸ = 𝟏𝟐𝟏 𝒙 = Β± 𝟏𝟐𝟏

𝒙 = ±𝟏𝟏 Try It! 2. Solve βˆ’5π‘₯3 + 80 = 0.

βˆ’πŸ“π’™πŸ

βˆ’πŸ“ =βˆ’πŸ–πŸŽβˆ’πŸ“

π’™πŸ = πŸπŸ” 𝒙 = Β± πŸπŸ” 𝒙 = Β±πŸ’

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BEAT THE TEST!

1. What is the smallest solution to the equation 2π‘₯3 + 17 = 179?

A βˆ’πŸ— B βˆ’3 C 3 D 9 Answer is A.

2. A rescuer on a helicopter that is 50 feet above the sea drops a lifebelt. The distance from the lifebelt to the sea can be modeled by the equation β„Ž(𝑑) = βˆ’16𝑑3 + 𝑠, where 𝑑 is the time, in seconds, after the lifebelt is dropped, and 𝑠 is the initial height, in feet, of the lifebelt above the sea.

How long will it take for the lifebelt to reach the sea? Round your answer to the nearest tenth of a second. βˆ’πŸπŸ”π’•πŸ + πŸ“πŸŽ = 𝟎 βˆ’πŸπŸ”π’•πŸ

βˆ’πŸπŸ” =βˆ’πŸ“πŸŽβˆ’πŸπŸ”

𝒕 = Β± πŸ“πŸŽπŸπŸ”

𝒕 = Β± πŸ“πŸ’πŸ or 𝒕 = ±𝟏. πŸ–

𝟏. πŸ–seconds

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Section 5 – Topic 7 Solving Quadratic Equations by Completing the Square

Sometimes you won’t be able to solve a quadratic equation by factoring. However, you can rewrite the quadratic equation so that you can complete the square to factor and solve. Let’s start by determining what number we can add to a quadratic expression to make it a perfect square trinomial.

What value could be added to the quadratic expression to make it a perfect square trinomial? π‘₯3 + 6π‘₯ + πŸ— π‘₯3 + 8π‘₯ + 3 + πŸπŸ‘ π‘₯3 βˆ’ 22π‘₯ βˆ’ 71 +192 Let’s see how this can be used to solve quadratic equations.

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Recall how we factored perfect square trinomials. If π‘Ž3 + 2π‘Žπ‘ + 𝑏3 is a perfect square trinomial, then π‘Ž3 + 2π‘Žπ‘ + 𝑏3 = π‘Ž + 𝑏 3 and π‘Ž3 βˆ’ 2π‘Žπ‘ + 𝑏3 = π‘Ž βˆ’ 𝑏 3. Solve 𝑓 π‘₯ = π‘Žπ‘₯3 + 𝑏π‘₯ + 𝑐 by completing the square. Step 1: Group π‘Žπ‘₯3and 𝑏π‘₯ together.

𝑓 π‘₯ = (π‘Žπ‘₯3 + 𝑏π‘₯ +___) + 𝑐 Step 2: If π‘Ž β‰  1, then factor out π‘Ž.

𝑓 π‘₯ = π‘Ž(π‘₯3 +π‘π‘Ž π‘₯ + ______) + 𝑐

Step 3: Divide j

k by two and square the result. Add that

number to the grouped terms. Subtract the product of that number and π‘Ž from 𝑐 so that you have not changed the equation.

𝑓 π‘₯ = π‘Ž(π‘₯3 + j

kπ‘₯ + j

l

^kl) + 𝑐 βˆ’ jl

^k

Step 4: Factor the trinomial.

𝑓 π‘₯ = π‘Ž π‘₯ +𝑏2π‘Ž

3

+4π‘Žπ‘ βˆ’ 𝑏3

4π‘Ž Step 5: This is vertex form. Now we can solve the equation by

setting the function equal to zero, moving the constant to the opposite side, and taking the square root of both sides.

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Let’s Practice! 1. Consider the following quadratic expression 2π‘₯3 βˆ’ 8π‘₯ + 5.

a. Complete the square to write the quadratic expression in vertex form. 𝟐(𝒙 βˆ’ 𝟐)𝟐 βˆ’ πŸ‘

b. If the expression represents a function, find the solutions to the quadratic function.

𝒙 β‰ˆ πŸ‘. 𝟐𝟐 or 𝒙 β‰ˆ 𝟎. πŸ•πŸ–

Try It! 2. Consider the quadratic expression 3π‘₯3 + 12π‘₯ + 31.

a. Complete the square to write the quadratic expression in vertex form. πŸ‘(𝒙 + 𝟐)𝟐 + πŸπŸ—

b. If the expression represents a function, find the solutions to the quadratic function. No Solution. This graph would not cross the 𝒙 βˆ’ axis.

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BEAT THE TEST! 1. The equations shown below are steps to solve the function

𝑔 π‘₯ = 2π‘₯3 + 24π‘₯ βˆ’ 29 by completing the square.

A. 𝑔 π‘₯ = 2 π‘₯3 + 12π‘₯ + 36 βˆ’ 29 βˆ’ 72

B. π‘₯ + 6 = Β± 50.5

C. 𝑔 π‘₯ = 2(π‘₯3 + 12π‘₯ + ____) βˆ’ 29

D. π‘₯ = βˆ’6 Β± 50.5

E. 2 π‘₯ + 6 3 βˆ’ 101 = 0

F. π‘₯ + 6 3 = 50.5

G. π‘₯ + 6 3 = Β± 50.5

H. 𝑔 π‘₯ = 2 π‘₯ + 6 3 βˆ’ 101

Place the equations in the correct order by writing the letter corresponding to each step in the boxes below.

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Step 1 Step 2 Step 3 Step 4

Step 8 Step 7 Step 6 Step 5

C A H E

D B G F

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Section 5 – Topic 8 Deriving the Quadratic Formula

We can use the process of completing the square to derive a formula to solve any quadratic equation. Consider the quadratic equation, 𝑓 π‘₯ = π‘Žπ‘₯3 + 𝑏π‘₯ + 𝑐, where π‘Ž β‰  0. Recall our steps for completing the square as a method for solving 𝑓(π‘₯). Step 1: Group π‘Žπ‘₯3and 𝑏π‘₯ together. 𝒇 𝒙 = π’‚π’™πŸ + 𝒃𝒙 + + 𝒄 Step 2: If π‘Ž β‰  1, then factor out π‘Ž.

𝒇 𝒙 = 𝒂 π’™πŸ +𝒃𝒂𝒙 +

π’ƒπŸ

πŸ’π’‚πŸ + 𝒄 βˆ’π’ƒπŸ

πŸ’π’‚πŸ

Step 3: Divide j

k by two and square the result. Add that

number to the grouped terms. Subtract the product of that number and π‘Ž from 𝑐 so that you have not changed the equation.

πŸ’π’‚π’„ βˆ’ π’ƒπŸ

πŸ’π’‚ Step 4: Factor the trinomial.

𝒇 𝒙 = 𝒂 𝒙 +π’ƒπŸπ’‚

𝟐

+πŸ’π’‚π’„ βˆ’ π’ƒπŸ

πŸ’π’‚

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Step 5: Solve the equation.

𝒙 +π’ƒπŸπ’‚

𝟐

=π’ƒπŸ βˆ’ πŸ’π’‚π’„πŸ’π’‚πŸ

𝒙 +π’ƒπŸπ’‚

𝟐

= Β±π’ƒπŸ βˆ’ πŸ’π’‚π’„πŸ’π’‚πŸ

𝒙 +π’ƒπŸπ’‚ =

Β± π’ƒπŸ βˆ’ πŸ’π’‚π’„πŸπ’‚

𝒙 = βˆ’π’ƒπŸπ’‚ Β±

π’ƒπŸ βˆ’ πŸ’π’‚π’„πŸπ’‚

𝒙 =βˆ’π’ƒ Β± π’ƒπŸ βˆ’ πŸ’π’‚π’„

πŸπ’‚

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BEAT THE TEST! 1. Complete the missing steps in the derivation of the

quadratic formula:

𝑓 π‘₯ = π‘Žπ‘₯3 + 𝑏π‘₯ + 𝑐

𝑓 π‘₯ = π‘Ž π‘₯3 + π‘π‘Ž π‘₯ +

𝑏2

4π‘Ž2 + 𝑐 βˆ’ 𝑏24π‘Ž

𝑓 π‘₯ = π‘Ž π‘₯ + 𝑏2π‘Ž

3+ 4π‘Žπ‘βˆ’π‘2

4π‘Ž

π‘Ž π‘₯ +𝑏2π‘Ž

3

+4π‘Žπ‘ βˆ’ 𝑏2

4π‘Ž = 0

π‘₯ + 𝑏2π‘Ž

3= 𝑏2βˆ’4π‘Žπ‘

4π‘Ž2

π‘₯ + 𝑏2π‘Ž = Β± 𝑏2βˆ’4π‘Žπ‘

2π‘Ž

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𝒇(𝒙) = nπ’‚π’™πŸ + 𝒃𝒙 +o + 𝒄+ 𝒄

pq𝒙 + π’ƒπŸπ’‚r𝟐= Β±π’ƒπŸsπŸ’π’‚π’„

πŸ’π’‚πŸ

𝒙 =βˆ’π’ƒ Β± βˆšπ’ƒπŸ βˆ’ πŸ’π’‚π’„

πŸπ’‚

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Section 5 – Topic 9 Solving Quadratic Equations Using the Quadratic

Formula For any quadratic equation π‘Žπ‘₯3 + 𝑏π‘₯ + 𝑐 = 0, where π‘Ž β‰  0,

π‘₯ =βˆ’π‘ Β± 𝑏3 βˆ’ 4π‘Žπ‘

2π‘Ž To use the quadratic formula:

Step 1: Set the quadratic equation equal to zero.

Step 2: Identify π‘Ž, 𝑏, and 𝑐.

Step 3: Substitute π‘Ž, 𝑏, and 𝑐 into the quadratic formula and evaluate to find the zeros.

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Let’s Practice! 1. Use the quadratic formula to solve π‘₯3 βˆ’ 4π‘₯ + 3 = 0.

𝒂 = 𝟏

𝒃 = βˆ’πŸ’ 𝒄 = πŸ‘

𝒙 =πŸ’Β± βˆ’πŸ’ πŸβˆ’πŸ’(𝟏)(πŸ‘)

𝟐(𝟏)

𝒙 = πŸ’Β± πŸπŸ”βˆ’πŸπŸπŸ(𝟏)

𝒙 =πŸ’ Β± πŸ’

𝟐

𝒙 =πŸ’ Β± 𝟐𝟐

𝒙 = πŸ‘ or 𝒙 = 𝟏 πŸ‘, 𝟏

2. Consider the graph of the quadratic equation 𝑦 = π‘₯3 βˆ’ 4π‘₯ + 3.

Does the graph verify the solutions we found using the quadratic formula? Yes, the 𝒙-intercepts are 𝟏 and πŸ‘.

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3. Use the quadratic formula to solve 2𝑀3 + 𝑀 = 5.

𝒂 = 𝟐 𝒃 = 𝟏 𝒄 = βˆ’πŸ“

π’˜ =βˆ’πŸΒ± 𝟏 πŸβˆ’πŸ’(𝟐)(βˆ’πŸ“)

𝟐(𝟐)

π’˜ = βˆ’πŸΒ± 𝟏+πŸ’πŸŽπŸ’

π’˜ =βˆ’πŸ Β± πŸ’πŸ

πŸ’ π’˜ β‰ˆ 𝟏. πŸ‘πŸ“ or π’˜ β‰ˆ βˆ’πŸ. πŸ–πŸ“ {βˆ’πŸ. πŸ–πŸ“, 𝟏. πŸ‘πŸ“}

Try It! 4. Use the quadratic formula to solve 3π‘ž3 βˆ’ 11 = 20π‘ž.

𝒂 = πŸ‘ 𝒃 = βˆ’πŸπŸŽ 𝒄 = βˆ’πŸπŸ

𝒒 =𝟐𝟎± βˆ’πŸπŸŽ πŸβˆ’πŸ’(πŸ‘)(βˆ’πŸπŸ)

𝟐(πŸ‘)

𝒒 = 𝟐𝟎± πŸ’πŸŽπŸŽ+πŸπŸ‘πŸπŸ”

𝒒 =𝟐𝟎 Β± πŸ“πŸ‘πŸ

πŸ” 𝒒 β‰ˆ πŸ•. πŸπŸ– or 𝒒 β‰ˆ βˆ’πŸŽ. πŸ“ {βˆ’πŸŽ. πŸ“, πŸ•. πŸπŸ–}

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BEAT THE TEST!

1. Your neighbor’s garden measures 12 meters by 16 meters. He plans to install a pedestrian pathway all around it, increasing the total area to 285 square meters. The new area can be represented by 4𝑀3 + 56𝑀 + 192. Use the quadratic formula to find the width, 𝑀, of the pathway.

Part A: Write an equation that can be used to solve for

the width of the pathway. πŸ’π’˜πŸ + πŸ“πŸ”π’˜ + πŸπŸ—πŸ = πŸπŸ–πŸ“

πŸ’π’˜πŸ + πŸ“πŸ”π’˜ βˆ’ πŸ—πŸ‘ = 𝟎

Part B: Use the quadratic formula to solve for the width of the pathway.

𝒂 = πŸ’ 𝒃 = πŸ“πŸ” 𝒄 = βˆ’πŸ—πŸ‘

π’˜ =βˆ’πŸ“πŸ”Β± πŸ“πŸ” πŸβˆ’πŸ’(πŸ’)(βˆ’πŸ—πŸ‘)

𝟐(πŸ’)

π’˜ = βˆ’πŸ“πŸ”Β± πŸ‘πŸπŸ‘πŸ”+πŸπŸ’πŸ–πŸ–πŸ–

π’˜ =βˆ’πŸ“πŸ” Β± πŸ’πŸ”πŸπŸ’

πŸ– π’˜ = 𝟏. πŸ“ or π’˜ = βˆ’πŸπŸ“. πŸ“

The width of the path is approximately 𝟏. πŸ“ meters.

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Section 5 – Topic 10 Quadratic Functions in Action

Let’s consider solving real-world situations that involve quadratic functions. Consider an object being launched into the air. We compare the height versus time elapsed. From what height is the object launched?

πŸ‘meters Once the object is launched, how long does it take to reach its maximum height?

𝟐. πŸ“ seconds

What is the maximum height? πŸ“meters Once the object is launched, how long does it take for it to hit the ground? πŸ”. πŸ“ seconds Once the object is launched, when does it return to a height of three meters? πŸ“seconds

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Question How to Answer it

1. From what height is the object launched?

This is the 𝑦-intercept. In the standard form, π‘Žπ‘₯3 + 𝑏π‘₯ + 𝑐, 𝑐is the 𝑦-intercept.

2. How long does it take for the object to reach its maximum height?

This is the π‘₯-coordinate of the vertex, π‘₯ = sj

3k, where values of π‘Ž

and 𝑏 come from the standard form of a quadratic equation. π‘₯ = sj3k is also the equation that

represents the axis of symmetry.

3. What is the maximum height?

This is the 𝑦-coordinate of the vertex. Substitute the π‘₯-coordinate from the step above and evaluate to find 𝑦. In vertex form, the height is π‘˜ and the vertex is (β„Ž, π‘˜).

4. How long does it take for the object to hit the ground?

The π‘₯-intercept(s) are the solution(s), or zero(s), of the quadratic function. Solve by factoring, using the quadratic formula, or by completing the square. In a graph, look at the π‘₯-intercept(s).

5. When does the object return to a height of three meters?

In function 𝐻 𝑑 = π‘Žπ‘‘3 + 𝑏𝑑 + 𝑐, if height is given, then substitute the value for 𝐻(𝑑). If time is given, then substitute for 𝑑.

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Let’s Practice! 1. Ferdinand is playing golf. He hits a shot off the tee box that

has a height modeled by the function β„Ž 𝑑 = βˆ’16𝑑3 + 80𝑑, where β„Ž(𝑑) is the height of the ball, in feet, and 𝑑 is the time in seconds it has been in the air. The graph that models the golf ball’s height over time is shown below.

a. When does the ball reach its maximum height?

𝟐. πŸ“ seconds

b. What is the maximum height of the ball?

𝟏𝟎𝟎feet

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c. What is the height of the ball at 3 seconds? When will the ball reach that same height again?

About πŸ—πŸ“ feet, About 𝟐 seconds

Try It! 2. Recall exercise 1.

a. When is the ball 65 feet in the air? Explain.

After about 𝟏 second and again after about πŸ’ seconds

b. How long does it take until the golf ball hits the ground? πŸ“ seconds

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BEAT THE TEST!

1. A neighborhood threw a fireworks celebration for the 4th of July. A bottle rocket was launched upward from the ground with an initial velocity of 160 feet per second. The formula for vertical motion of an object is

β„Ž 𝑑 = 0.5π‘Žπ‘‘3 + 𝑣𝑑 + 𝑠, where the gravitational constant, π‘Ž, is βˆ’32 feet per square second, 𝑣 is the initial velocity, 𝑠 is the initial height, and β„Ž(𝑑) is the height in feet modeled as a function of time, 𝑑.

Part A: What function describes the height, β„Ž, of the

bottle rocket after 𝑑 seconds have elapsed?

𝒉 𝒕 = βˆ’πŸπŸ”π’•πŸ + πŸπŸ”πŸŽπ’• Part B: What was the maximum height of the bottle

rocket? Time to reach max. height: 𝒕 = βˆ’ πŸπŸ”πŸŽ

𝟐 βˆ’πŸπŸ” = πŸ“ seconds

Maximum height: 𝒉 πŸ“ = βˆ’πŸπŸ” πŸ“ 𝟐 + πŸπŸ”πŸŽ πŸ“ = πŸ’πŸŽπŸŽ feet

Test Yourself! Practice Tool

Great job! You have reached the end of this section. Now it’s time to try the β€œTest Yourself! Practice Tool,” where you can practice all the skills and concepts you learned in this section. Log in to Algebra Nation and try out the β€œTest Yourself! Practice Tool” so you can see how well you know these topics!

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