ORCCA Solving Quadratic Equations by Factoring

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Section 10.7 Solving Quadratic Equations by Factoring

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Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-solve-quadratic-equations-using-zero-product-principle" missing or not unique]
[cross-reference to target(s) "ccog-check-solutions-to-equations" missing or not unique]
[cross-reference to target(s) "ccog-solve-applications-involving-quadratic-and-rational-equations" missing or not unique]

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Subsection 10.7.1 Introduction

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We have learned how to factor trinomials like

x^{2} + 5 x + 6

into

(x + 2) (x + 3) .

This skill can be used to solve an equation like

x^{2} + 5 x + 6 = 0,

which is a quadratic equation. Note that we solved equations like this in Chapter 7, but here we will use factoring, a new method.

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Definition 10.7.1. Quadratic Equation.

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A quadratic equation is is an equation in the form

a x^{2} + b x + c = 0

with

a \neq 0 .

We also consider equations such as

x^{2} = x + 3

and

5 x^{2} + 3 = (x + 1)^{2} + (x + 1) (x - 3)

to be quadratic equations, because we can expand any multiplication, add or subtract terms from both sides, and combine like terms to get the form

a x^{2} + b x + c = 0 .

The form

a x^{2} + b x + c = 0

is called the standard form of a quadratic equation.

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Figure 10.7.2. Alternative Video Lesson

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Before we begin exploring the method of solving quadratic equations by factoring, we’ll identify what types of equations are quadratic and which are not.

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Checkpoint 10.7.3.

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Identify which of the items are quadratic equations.

The equation $2 x^{2} + 5 x = 7$
- ?
- is
- is not
a quadratic equation.
The equation $5 - 2 x = 3$
- ?
- is
- is not
a quadratic equation.
The equation $15 - x^{3} = 3 x^{2} + 9 x$
- ?
- is
- is not
a quadratic equation.
The equation $(x + 3) (x - 4) = 0$
- ?
- is
- is not
a quadratic equation.
The equation $x (x + 1) (x - 1) = 0$
- ?
- is
- is not
a quadratic equation.
The expression $x^{2} - 5 x + 6$
- ?
- is
- is not
a quadratic equation.
The equation $(2 x - 3) (x + 5) = 12$
- ?
- is
- is not
a quadratic equation.

Explanation.

The equation $2 x^{2} + 5 x = 7$ is a quadratic equation. To write it in standard form, simply subtract $7$ from both sides.
The equation $5 - 2 x = 3$ is not quadratic. It is a linear equation.
The equation $15 + x^{3} = 3 x^{2} + 9 x$ is not a quadratic equation because of the $x^{3}$ term.
The equation $(x + 3) (x - 4) = 0$ is a quadratic equation. If we expand the left-hand side of the equation, we would get something in standard form.
The equation $x (x + 1) (x - 1) = 0$ is not a quadratic equation. If we expanded the left-hand side of the equation, we would have an expression with an $x^{3}$ term, which automatically makes it not quadratic.
The expression $x^{2} - 5 x + 6$ is not a quadratic equation; it’s not an equation at all. Instead, this is a quadratic expression.
The equation $(2 x - 3) (x + 5) = 12$ is a quadratic equation. Multiplying out the left-hand side, and subtracting $12$ form both sides, we would have a quadratic equation in standard form.

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Now we’ll look at an application that demonstrates the need and method for solving a quadratic equation by factoring.

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Nita is in a physics class that launches a tennis ball from a rooftop

80

feet above the ground. They fire it directly upward at a speed of

64

feet per second and measure the time it takes for the ball to hit the ground below. We can model the height of the tennis ball,

h,

in feet, with the quadratic equation

h = - 16 t^{2} + 64 t + 80,

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where

t

represents the time in seconds after the launch. Using the model we can predict when the ball will hit the ground.

A drawing of a tall building and a tennis ball being launched upward; the path of the ball is an arc that goes upward and then downward to the ground — Figure 10.7.4. A Diagram of the Ball Thrown from the Roof

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The ground has a height of

0,

h = 0 .

We will substitute

0

for

h

in the equation and we have

- 16 t^{2} + 64 t + 80 = 0

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We need to solve this quadratic equation for

t

to find when the ball will hit the ground.

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The key strategy for solving a linear equation is to separate the variable terms from the constant terms on either side of the equal sign. It turns out that this same method will not work for quadratic equations. Fortunately, we already have spent a good amount of time discussing a method that will work: factoring. If we can factor the polynomial on the left-hand side, we will be on the home stretch to solving the whole equation.

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We will look for a common factor first, and see that we can factor out

- 16 .

Then we can finish factoring the trinomial:

\begin{aligned} - 16 t^{2} + 64 t + 80 & = 0 \\ - 16 (t^{2} - 4 t - 5) & = 0 \\ - 16 (t + 1) (t - 5) & = 0 \end{aligned}

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In order to finish solving the equation, we need to understand the following property. This property explains why it was incredibly important to

n o t

move the

80

in our example over to the other side of the equation before trying to factor.

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Fact 10.7.5. Zero Product Property.

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If the product of two or more numbers is equal to zero, then at least one of the numbers must be zero.

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One way to understand this property is to think about the equation

a \cdot b = 0 .

Maybe

b = 0,

because that would certainly cause the equation to be true. But suppose that

b \neq 0 .

Then it is safe to divide both sides by

b,

and the resulting equation says that

a = 0 .

So no matter what, either

a = 0

b = 0 .

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To understand this property more, let’s look at a few products:

\begin{aligned} 4 \cdot 7 & = 28 & 4 \cdot 0 & = 0 & 4 \cdot 7 \cdot 3 & = 84 \\ 0 \cdot 7 & = 0 & - 4 \cdot 0 & = 0 & 4 \cdot 0 \cdot 3 & = 0 \end{aligned}

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When none of the factors are

0,

the result is never

0 .

The only way to get a product of

0

is when one of the factors is

0 .

This property is unique to the number

0

and can be used no matter how many numbers are multiplied together.

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Now we can see the value of factoring. We have three factors in our equation

- 16 (t + 1) (t - 5) = 0 .

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The first factor is the number

- 16 .

The second and third factors,

t + 1

and

t - 5,

are expressions that represent numbers. Since the product of the three factors is equal to

0,

one of the factors must be zero.

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Since

- 16

is not

0,

either

t + 1

t - 5

must be

0 .

This gives us two equations to solve:

\begin{aligned} t + 1 & = 0 & or & t - 5 & = 0 \\ t + 1 - 1 & = 0 - 1 & or & t - 5 + 5 & = 0 + 5 \\ t & = - 1 & or & t & = 5 \end{aligned}

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We have found two solutions,

- 1

and

5 .

A quadratic expression will have at most two linear factors, not including any constants, so it can have up to two solutions.

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Let’s check each of our two solutions

- 1

and

5 :

\begin{aligned} - 16 t^{2} + 64 t + 80 & = 0 & - 16 t^{2} + 64 t + 80 & = 0 \\ - 16 (- 1)^{2} + 64 (- 1) + 80 & \overset{?}{=} 0 & - 16 (5)^{2} + 64 (5) + 80 & \overset{?}{=} 0 \\ - 16 (1) - 64 + 80 & \overset{?}{=} 0 & - 16 (25) + 320 + 80 & \overset{?}{=} 0 \\ - 16 - 64 + 80 & \overset{?}{=} 0 & - 400 + 320 + 80 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 & 0 & \overset{✓}{=} 0 \end{aligned}

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We have verified our solutions. While there are two solutions to the equation, the solution

- 1

is not relevant to this physics model because it is a negative time which would tell us something about the ball’s height before it was launched. The solution

5

does make sense. According to the model, the tennis ball will hit the ground

5

seconds after it is launched.

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Subsection 10.7.2 Further Examples

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We’ll now look at further examples of solving quadratic equations by factoring. The general process is outlined here:

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Process 10.7.6. Solving Quadratic Equations by Factoring.

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Simplify: 🔗
Simplify the equation using distribution and by combining like terms.
Isolate: 🔗
Move all terms onto one side of the equation so that the other side has $0 .$
Factor: 🔗
Factor the quadratic expression.
Apply the Zero Product Property: 🔗
Apply the Zero Product Property.
Solve: 🔗
Solve the equation(s) that result after the zero product property was applied.

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Example 10.7.7.

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Solve

x^{2} - 5 x - 14 = 0

by factoring.

Explanation.

\begin{aligned} x^{2} - 5 x - 14 & = 0 \\ (x - 7) (x + 2) & = 0 \end{aligned}

\begin{aligned} x - 7 & = 0 & or & x + 2 & = 0 \\ x - 7 + 7 & = 0 + 7 & or & x + 2 - 2 & = 0 - 2 \\ x & = 7 & or & x & = - 2 \end{aligned}

The solutions are

- 2

and

7,

so the solution set is written as

{- 2, 7} .

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Example 10.7.8.

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Solve

x^{2} - 5 x - 14 = 0

by using graphing technology.

Explanation.

We have already solved the equation

x^{2} - 5 x - 14 = 0

by factoring in Example 7, and now we can analyze the significance of the solutions graphically. What will

- 2

and

7

mean on the graph?

To solve this equation graphically, we first make a graph of

y = x^{2} - 5 x - 14

and of

y = 0 .

Both of these graphs are shown in Figure 9 in an appropriate window.

Figure 10.7.9. A graph of

y = x^{2} - 5 x - 14

and

y = 0 .

From the graph provided we can see that the solutions (the

x

-values where the graphs intersect) are

- 2

and

7 .

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If the two factors of a polynomial happen to be the same, the equation will only have one solution. Let’s look at an example of that.

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Example 10.7.10. A Quadratic Equation with Only One Solution.

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Solve

x^{2} - 10 x + 25 = 0

by factoring.

Explanation.

\begin{aligned} x^{2} - 10 x + 25 & = 0 \\ (x - 5) (x - 5) & = 0 \\ (x - 5)^{2} & = 0 \\ x - 5 & = 0 \\ x - 5 + 5 & = 0 + 5 \\ x & = 5 \end{aligned}

The solution is

5,

so the solution set is written as

{5} .

While we are examining this problem, let’s compare the algebraic solution to a graphical solution.

To solve this equation graphically, we first make a graph of

y = x^{2} - 10 x + 25

and of

y = 0 .

Both of these graphs are shown in Figure 11 in an appropriate window.

Figure 10.7.11. A graph of

y = x^{2} - 10 x + 25

and

y = 0 .

From the graph provided, we can see that the reason that the equation

x^{2} - 10 x + 25 = 0

has only one solution is that the parabola

y = x^{2} - 10 x + 25

touches the line

y = 0

only once. So again, the solution is

5 .

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Example 10.7.12. Factor Out a Common Factor.

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Solve

5 x^{2} + 55 x + 120 = 0

by factoring.

Explanation.

Note that the terms are all divisible by

5,

so we can factor that out to start.

\begin{aligned} 5 x^{2} + 55 x + 120 & = 0 \\ 5 (x^{2} + 11 x + 24) & = 0 \\ 5 (x + 8) (x + 3) & = 0 \end{aligned}

\begin{aligned} x + 8 & = 0 & or & x + 3 & = 0 \\ x & = - 8 & or & x & = - 3 \end{aligned}

The solution set is

{- 8, - 3} .

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Example 10.7.13. Factoring Using the AC Method.

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Solve

3 x^{2} - 7 x + 2 = 0

by factoring.

Explanation.

Recall that we multiply

3 \cdot 2 = 6

and find a factor pair that multiplies to

6

and adds to

- 7 .

The factors are

- 6

and

- 1 .

We use the two factors to replace the middle term with

- 6 x

and

- x .

\begin{aligned} 3 x^{2} - 7 x + 2 & = 0 \\ 3 x^{2} - 6 x - x + 2 & = 0 \\ (3 x^{2} - 6 x) + (- x + 2) & = 0 \\ 3 x (x - 2) - 1 (x - 2) & = 0 \\ (3 x - 1) (x - 2) & = 0 \end{aligned}

\begin{aligned} 3 x - 1 & = 0 & or & x - 2 & = 0 \\ 3 x & = 1 & or & x & = 2 \\ x & = \frac{1}{3} & or & x & = 2 \end{aligned}

The solution set is

{\frac{1}{3}, 2} .

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So far the equations have been written in standard form, which is

a x^{2} + b x + c = 0

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If an equation is not given in standard form then we must rearrange it in order to use the Zero Product Property.

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Example 10.7.14. Writing in Standard Form.

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Solve

x^{2} - 10 x = 24

by factoring.

Explanation.

There is nothing like the Zero Product Property for the number

24 .

We must have a

0

on one side of the equation to solve quadratic equations using factoring.

\begin{aligned} x^{2} - 10 x & = 24 \\ x^{2} - 10 x - 24 & = 24 - 24 \\ x^{2} - 10 x - 24 & = 0 \\ (x - 12) (x + 2) & = 0 \end{aligned}

\begin{aligned} x - 12 & = 0 & or & x + 2 & = 0 \\ x & = 12 & or & x = & - 2 \end{aligned}

The solution set is

{- 2, 12} .

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Example 10.7.15. Writing in Standard Form.

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Solve

(x + 4) (x - 3) = 18

by factoring.

Explanation.

Again, there is nothing like the Zero Product Property for a number like

18 .

We must expand the left side and subtract

18

from both sides.

\begin{aligned} (x + 4) (x - 3) & = 18 \\ x^{2} + x - 12 & = 18 \\ x^{2} + x - 12 - 18 & = 18 - 18 \\ x^{2} + x - 30 & = 0 \\ (x + 6) (x - 5) & = 0 \end{aligned}

\begin{aligned} x + 6 & = 0 & or & x - 5 & = 0 \\ x & = - 6 & or & x & = 5 \end{aligned}

The solution set is

{- 6, 5} .

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Example 10.7.16. A Quadratic Equation with No Constant Term.

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Solve

2 x^{2} = 5 x

by factoring.

Explanation.

It may be tempting to divide both sides of the equation by

x .

But

x

is a variable, and for all we know, maybe

x = 0 .

So it is not safe to divide by

x .

As a general rule, never divide an equation by a variable in the solving process. Instead, we will put the equation in standard form.

\begin{aligned} 2 x^{2} & = 5 x \\ 2 x^{2} - 5 x & = 5 x - 5 x \\ 2 x^{2} - 5 x & = 0 \end{aligned}

We can factor out

x .

\begin{aligned} x (2 x - 5) & = 0 \end{aligned}

\begin{aligned} x & = 0 & or & 2 x - 5 & = 0 \\ x & = 0 & or & 2 x & = 5 \\ x & = 0 & or & x & = \frac{5}{2} \end{aligned}

The solution set is

{0, \frac{5}{2}} .

In general, if a quadratic equation does not have a constant term, then

0

will be one of the solutions.

While we are examining this problem, let’s compare the algebraic solution to a graphical solution.

To solve this equation graphically, we first make a graph of

y = 2 x^{2}

and of

y = 5 x .

Both of these graphs are shown in Figure 17 in an appropriate window.

Figure 10.7.17. A graph of

y = 2 x^{2}

and

y = 5 x .

From the graph provided, we can see that the equation

2 x^{2} = 5 x

has two solutions because the graph of

y = 2 x^{2}

crosses the graph of

y = 5 x

twice. Those solutions, the

x

-values where the graphs cross, appear to be

0

and

\frac{5}{2} = 2.5 .

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Example 10.7.18. Factoring a Special Polynomial.

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Solve

x^{2} = 9

by factoring.

Explanation.

We can put the equation in standard form and use factoring. In this case, we find a difference of squares.

\begin{aligned} x^{2} & = 9 \\ x^{2} - 9 & = 0 \\ (x + 3) (x - 3) & = 0 \end{aligned}

\begin{aligned} x + 3 & = 0 & or & x - 3 & = 0 \\ x & = - 3 & or & x & = 3 \end{aligned}

The solution set is

{- 3, 3} .

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Example 10.7.19. Solving an Equation with a Higher Degree.

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Solve

2 x^{3} - 10 x^{2} - 28 x = 0

by factoring.

Explanation.

Although this equation is not quadratic, it does factor so we can solve it by factoring.

\begin{aligned} 2 x^{3} - 10 x^{2} - 28 x & = 0 \\ 2 x (x^{2} - 5 x - 14) & = 0 \\ 2 x (x - 7) (x + 2) & = 0 \end{aligned}

\begin{aligned} 2 x & = 0 & or & x - 7 & = 0 & or & x + 2 & = 0 \\ x & = 0 & or & x & = 7 & or & x & = - 2 \end{aligned}

The solution set is

{- 2, 0, 7} .

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Subsection 10.7.3 Applications

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Example 10.7.20. Kicking it on Mars.

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Some time in the recent past, Filip traveled to Mars for a vacation with his kids, Henrik and Karina, who wanted to kick a soccer ball around in the comparatively reduced gravity. Karina stood at point

K

and kicked the ball over her dad standing at point

F

to Henrik standing at point

H .

The height of the ball off the ground,

h

in feet, can be modeled by the equation

h = - 0.01 (x^{2} - 70 x - 1800),

where

x

is how far to the right the ball is from Filip. Note that distances to the left of Filip will be negative.

Find out how high the ball was above the ground when it passed over Filip’s head.
Find the distance from Karina to Henrik.

a coordinate plane with a dot for Karina, Filip and Henrik; There is an arc from Karina to Henrik that goes over Filip — Figure 10.7.21. A Soccer Kick on Mars

Explanation.

The ball was neither left nor right of Filip when it went over him, so $x = 0 .$ Plugging that value into our equation for $x,$

$\begin{aligned} h & = - 0.01 (0^{2} - 70 (0) - 1800) \\ = - 0.01 (- 1800) \\ = 18 \end{aligned}$

It seems that the soccer ball was $18$ feet above the ground when it flew over Filip.
The distance from Karina to Henrik is the same as the distance from point $K$ to point $H .$ These are the horizontal intercepts of the graph of the given formula: $h = - 0.01 (x^{2} - 70 x - 1800) .$ To find the horizontal intercepts, set $h = 0$ and solve for $x .$

$\begin{array}{r} 0 = - 0.01 (x^{2} - 70 x - 1800) \end{array}$
Note that we can divide by $- 0.01$ on both sides of the equation to simplify.
$\begin{aligned} 0 & = x^{2} - 70 x - 1800 \\ 0 & = (x - 90) (x + 20) \end{aligned}$
So, either:
$\begin{aligned} x - 90 & = 0 & or & x + 20 & = 0 \\ x & = 90 & or & x & = - 20 \end{aligned}$

Since the $x$ -values are how far right or left the points are from Filip, Karina is standing $20$ feet left of Filip and Henrik is standing $90$ feet right of Filip. Thus, the two kids are $110$ feet apart.

It is worth noting that if this same kick, with same initial force at the same angle, took place on Earth, the ball would have traveled less than

30

feet from Karina before landing!

The previous coordinate plane with a scale added so that Filip is standing at the origin; the scale on the x-axis is 20 feet and the scale on the y-axis is 10 feet; there is an arc for the soccer ball on Earth that would travel 30 feet; the arc for the ball kicked on the moon would travel 110 feet — Figure 10.7.22. A Soccer Kick on Mars and the Same Kick on Earth

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Example 10.7.23. An Area Application.

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Rajesh has a hot tub and he wants to build a deck around it. The hot tub is 7 ft by 5 ft and it is covered by a roof that is 99 ft². How wide can he make the deck so that it will be covered by the roof?

Explanation.

We will define

x

to represent the width of the deck (in feet). Here is a diagram to help us understand the scenario.

A diagram with a 7x5 rectangle for the hot tub; there is a larger rectangle around the hot tub for the deck, which extends x feet on each side of the hot tub — Figure 10.7.24. Diagram for the Deck

The overall length is

7 + 2 x

feet, because Rajesh is adding

x

feet on each side. Similarly, the overall width is

5 + 2 x

feet.

The formula for the area of a rectangle is

area = length \cdot width .

Since the total area of the roof is 99 ft², we can write and solve the equation:

\begin{aligned} (7 + 2 x) (5 + 2 x) = 99 \\ 4 x^{2} + 24 x + 35 = 99 \\ 4 x^{2} + 24 x + 35 - 99 & = 99 - 99 \\ 4 x^{2} + 24 x - 64 & = 0 \\ 4 (x^{2} + 6 x - 16) & = 0 \\ 4 (x + 8) (x - 2) & = 0 \end{aligned}

\begin{aligned} x + 8 & = 0 & or & x - 2 & = 0 \\ x & = - 8 & or & x & = 2 \end{aligned}

Since a length cannot be negative, we take

x = 2

as the only applicable solution. Rajesh should make the deck 2 ft wide on each side to fit under the roof.

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Reading Questions 10.7.4 Reading Questions

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1.

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If you use factoring to solve a polynomial equation, what number should be the only thing on one side of the equation?

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2.

🔗

When you are trying to solve a quadratic equation where the leading coefficient is not

1

m and you want to use factoring, and you have moved all terms to one side so that the other side is

0,

what should you look for before trying anything else?

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The rectangle’s height is .

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The rectangle’s base is .

🔗

72.

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A rectangle’s base is

9 c m

longer than its height. The rectangle’s area is

136 c m^{2} .

Find this rectangle’s dimensions.

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The rectangle’s height is .

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The rectangle’s base is .

🔗

73.

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A rectangle’s base is

2 i n

shorter than three times its height. The rectangle’s area is

65 i n^{2} .

Find this rectangle’s dimensions.

🔗

The rectangle’s height is .

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The rectangle’s base is .

🔗

74.

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A rectangle’s base is

3 i n

shorter than twice its height. The rectangle’s area is

35 i n^{2} .

Find this rectangle’s dimensions.

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The rectangle’s height is .

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The rectangle’s base is .

🔗

75.

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There is a rectangular lot in the garden, with

9 f t

in length and

3 f t

in width. You plan to expand the lot by an equal length around its four sides, and make the area of the expanded rectangle

135 f t^{2} .

How long should you expand the original lot in four directions?

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You should expand the original lot by in four directions.

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76.

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There is a rectangular lot in the garden, with

6 f t

in length and

4 f t

in width. You plan to expand the lot by an equal length around its four sides, and make the area of the expanded rectangle

80 f t^{2} .

How long should you expand the original lot in four directions?

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You should expand the original lot by in four directions.

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Challenge.

🔗

77.

🔗

Give an example of a cubic equation that has three solutions: one solution is

x = 3,

the second solution is

x = - 9,

and the third solution is

x = \frac{2}{3} .

Write your equation in standard form.

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78.

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Solve for

x

in the equation

75 x^{20} - 3 x^{18} = 0 .

You have attempted 1 of 2 activities on this page.

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Section 10.7 Solving Quadratic Equations by Factoring

Objectives: PCC Course Content and Outcome Guide

Subsection 10.7.1 Introduction

Definition 10.7.1. Quadratic Equation.

Checkpoint 10.7.3.

Fact 10.7.5. Zero Product Property.

Subsection 10.7.2 Further Examples

Process 10.7.6. Solving Quadratic Equations by Factoring.

Example 10.7.7.

Example 10.7.8.

Example 10.7.10. A Quadratic Equation with Only One Solution.

Example 10.7.12. Factor Out a Common Factor.

Example 10.7.13. Factoring Using the AC Method.

Example 10.7.14. Writing in Standard Form.

Example 10.7.15. Writing in Standard Form.

Example 10.7.16. A Quadratic Equation with No Constant Term.

Example 10.7.18. Factoring a Special Polynomial.

Example 10.7.19. Solving an Equation with a Higher Degree.

Subsection 10.7.3 Applications

Example 10.7.20. Kicking it on Mars.

Example 10.7.23. An Area Application.

Reading Questions 10.7.4 Reading Questions

1.

2.

Exercises 10.7.5 Exercises

Warmup and Review.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Solve Quadratic Equations by Factoring.

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