Int-Alg Intercepts, Solutions, and Factors

Section 3.2 Intercepts, Solutions, and Factors

In the last section, we used extraction of roots to solve quadratic equations of the form

a (x - p)^{2} = q

But this technique will not work on quadratic equations that also include a linear term,

b x .

Recall that the most general type of quadratic equation looks like

🔗

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

Here is an example.

🔗

Suppose a baseball player pops up, that is, she hits the baseball straight up into the air. The height,

h,

of the baseball after

t

seconds is given by a formula from physics. This formula takes into account the initial speed of the ball (64 feet per second) and its height when it was hit (4 feet).

🔗

h = - 16 t^{2} + 64 t + 4

The graph of this equation is shown below.

🔗

$baseball popup$

We would like to know when the baseball was exactly 52 feet high. To find out, we must solve the equation

🔗

- 16 t^{2} + 64 t + 4 = 52

where we have substituted 52 for the height,

h .

We can use the graph to solve this equation, by finding points with

h

-coordinate 52. You can see that there are two such points, with

t

-coordinates 1 and 3, so the baseball is 52 feet high at 1 second, and again on the way down at 3 seconds.

🔗

Can we solve the equation algebraically? Not with the techniques we know, because there are two terms containing the variable

t,

and they cannot be combined. We will need a new method.

[TK]

To find this method, we are going to study the connection between:

🔗

the factors of $a x^{2} + b x + c,$
the solutions of the quadratic equation $a x^{2} + b x + c = 0,$ and
the $x$ -intercepts of the graph of $y = a x^{2} + b x + c = 0 .$

🔗

Subsection 3.2.1 Zero-Factor Principle

The method we will learn now is not like extraction of roots, or like solving linear equations, where we "undid" in reverse order each operation performed on the variable, like peeling an onion. This new method will seem less direct. It relies on applying a property of our number system.

🔗

Can you multiply two numbers together and obtain a product of zero? Only if one of the two numbers happens to be zero. (Try it yourself.)

🔗

Zero-Factor Principle.

The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols,

🔗

a b = 0 if and only if a = 0 or b = 0 (or both)

🔗

Checkpoint 3.2.1. QuickCheck 1.

Fill in the blanks.

🔗

If the sum of two numbers is zero, the numbers must be .
If a fraction equals zero, the must be zero.
If the product of two numbers is zero, one of the numbers must be .
If the divisor in a quotient is zero, the quotient is .

🔗

Here is the simplest possible application of the Zero-Factor Principle (ZFP): For what value(s) of

x

is the equation

3 x = 0

true? You could divide both sides by 3, but you can also see that the product

3 x

can equal zero only if one of its factors is zero, so

x

must be zero!

🔗

The ZFP is true even if the numbers

a

and

b

are represented by algebraic expressions, such as

x - 6

x + 2 .

For example, if

🔗

(x - 6) (x + 2) = 0

then it must be true that either

x - 6 = 0

x + 2 = 0.

This is how we can use the ZFP to solve quadratic equations.

🔗

Example 3.2.2.

Solve the equation

x^{2} - 4 x - 12 = 0

🔗

Solution.

We can factor the expression

x^{2} - 4 x - 12,

and write the equation as

[TK]

🔗

(x - 6) (x + 2) = 0

Now it is in the form

a b = 0,

with

a = x - 6

and

b = x + 2,

so the ZFP tells us that either

x - 6 = 0

x + 2 = 0 .

We solve each of these equations.

🔗

\begin{aligned} x - 6 = 0 & or x + 2 = 0 & Solve each equation. \\ x = 6 & or x = - 2 \end{aligned}

Once again we see that a quadratic equation has two solutions. You can check that both of these values satisfy the original equation.

🔗

Checkpoint 3.2.3. Practice 1.

Solve the equation

x^{2} - 11 x + 24 = 0

🔗

Answer.

x = 3

x = 8

🔗

Subsection 3.2.2 X-Intercepts of a Parabola

Recall that the

x

-intercept of a line is the point where

y = 0,

or where the line crosses the

x

-axis. We find the

x

-intercept by setting

y = 0

in the equation of the line, and solving for

x .

We can find the

x

-intercepts of a parabola the same way.

🔗

Example 3.2.4.

Find the

x

-intercepts of the graph of

y = x^{2} - 4 x - 12

🔗

Solution.

To find the

x

-intercepts of the graph, we set

y = 0

and solve the equation

🔗

0 = x^{2} - 4 x - 12

But this is the same equation we solved in the last Example, because

🔗

x^{2} - 4 x - 12 = (x - 6) (x + 2)

The solutions of that equation were

6

and

- 2,

so the

x

-intercepts of the graph are

(6, 0)

and

(- 2, 0) .

You can see this by graphing the equation on your calculator, as shown in the figure.

🔗

🔗

We can state a general result: The

x

-intercepts of the graph of

🔗

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

are the solutions of the equation

🔗

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

So we can always solve a quadratic equation to find the

x

-intercepts of a parabola (if there are any).

🔗

And we can use this relationship the other way round, too: If we know the

x

-intercepts of the graph of

y = a x^{2} + b x + c,

we also know the solutions of the equation

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

🔗

Checkpoint 3.2.5. Practice 2.

Use technology to graph the equation

🔗

\begin{matrix} y = (x - 3) (2 x + 3) \end{matrix}

and find the

x

-intercepts of the graph. Use your answers to solve the equation

🔗

\begin{matrix} (x - 3) (2 x + 3) = 0 \end{matrix}

Check your solutions by applying the ZFP.

🔗

Answer.

x = 3

x = \frac{- 3}{2}

🔗

Subsection 3.2.3 Solving Quadratic Equations by Factoring

Now we’ll consider some other quadratic equations. Before we apply the ZFP, we must write the equation so that one side is zero.

🔗

Example 3.2.6.

Solve

3 x (x + 1) = 2 x + 2

🔗

Solution.

First, we write the equation in standard form.

🔗

\begin{aligned} 3 x (x + 1) & = 2 x + 2 & Apply the distributive law to the left side. \\ 3 x^{2} + 3 x & = 2 x + 2 & Subtract 2 x + 2 from both sides. \\ 3 x^{2} + x - 2 & = 0 \end{aligned}

Now we factor the left side to obtain

🔗

\begin{aligned} (3 x - 2) (x + 1) & = 0 & Apply the zero-factor principle. \\ 3 x - 2 = 0 or x + 1 & = 0 & Solve each equation. \\ x = \frac{2}{3} or x & = - 1 \end{aligned}

The solutions are

\frac{2}{3}

and

- 1 .

🔗

Caution 3.2.7.

When we apply the zero-factor principle, one side of the equation must be zero. For example, to solve the equation

🔗

(x - 2) (x - 4) = 15

it is incorrect to set each factor equal to 15! (There are many ways that the product of two numbers can equal 15; it is not necessary that one of the numbers be 15.)

🔗

We must first simplify the left side and write the equation in standard form. (The correct solutions are

7

and

- 1;

check that you can find these solutions.)

🔗

We summarize the factoring method for solving quadratic equations as follows.

🔗

To Solve a Quadratic Equation by Factoring.

Write the equation in standard form.
Factor the left side of the equation.
Apply the zero-factor principle: Set each factor equal to zero.
Solve each equation. There are two solutions (which may be equal).

🔗

Checkpoint 3.2.8. Practice 3.

Solve by factoring:

(t - 3)^{2} = 3 (9 - t)

🔗

Hint: Begin by multiplying out each side of the equation.

🔗

Answer.

t = - 3

t = 6

🔗

Checkpoint 3.2.9. QuickCheck 2.

Which technique, extracting roots or factoring, is better-suited to each equation?

🔗

$4 x^{2} - 12 x = 0$
$6 (4 x - 1)^{2} = 18$
$(x + 4)^{2} = 16 x$
$9 x^{2} - 42 = 0$

🔗

Now we can use factoring to solve the opening problem in this section.

🔗

Example 3.2.10.

The height,

h,

of a baseball

t

seconds after being hit is given by

🔗

h = - 16 t^{2} + 64 t + 4

When will the baseball reach a height of 64 feet?

🔗

Solution.

We substitute

64

for

h

in the formula, and solve for

t .

🔗

\begin{aligned} - 16 t^{2} + 64 t + 4 & = 64 & Write the equation in standard form. \\ 16 t^{2} - 64 t + 60 & = 0 & Factor 4 from the left side. \\ 4 (4 t^{2} - 16 t + 15) & = 0 & Factor the quadratic expression. \\ 4 (2 t - 3) (2 t - 50) & = 0 & Set each variable factor equal to zero. \\ 2 t - 3 = 0 or 2 t - 5 & = 0 & Solve each equation. \\ t = \frac{3}{2} or t & = \frac{5}{2} \end{aligned}

There are two solutions. At

t = \frac{3}{2}

seconds, the ball reaches a height of 64 feet on the way up, and at

t = \frac{5}{2}

seconds, the ball is 64 feet high on its way down.

🔗

Caution 3.2.11.

In the Example above, the factor of 4 does not affect the solutions of the equation at all. You can understand why this is true by looking at some graphs. Use technology to graph the equation

🔗

y_{1} = x^{2} - 4 x + 3

in the window

🔗

\begin{matrix} Xmin = - 2 Ymin = - 5 \\ Xmax = 8 Ymax = 10 \end{matrix}

Notice that when

y = 0,

x = 1

x = 3 .

These two points are the

x

-intercepts of the graph. Now on the same window graph

🔗

y_{2} = 4 (x^{2} - 4 x + 3)

as shown below.

🔗

This graph has the same

x

-values when

y = 0 .

The factor of 4 makes the graph "skinnier," but does not change the location of the

x

-intercepts.

🔗

Checkpoint 3.2.12. Practice 4.

Solve by factoring $4 t - t^{2} = 0 .$
Solve by factoring $20 t - 5 t^{2} = 0 .$
Graph $y = 4 t - t^{2}$ and $y = 20 t - 5 t^{2}$ together in the window

$\begin{aligned} Xmin & = - 2 & Ymin & = - 20 \\ Xmax & = 6 & Ymax & = 25 \end{aligned}$

and locate the horizontal intercepts on each graph.

🔗

Answer.

t = 0

and

t = 4

🔗

Checkpoint 3.2.13. QuickCheck 3.

Match each equation with its solutions.

🔗

$3 t (t - 2) = 0$
$t^{2} - t = 2$
$3 t^{2} = 12$
$t (t - 3) = 2 (t - 3)$

🔗

$- 1, 2$
$0, 2$
$2, 3$
$- 2, 2$

🔗

Subsection 3.2.4 An Application

Here is another example of how quadratic equations arise in applications.

🔗

Example 3.2.14.

The size of a rectangular computer monitor screen is taken to be the length of its diagonal. If the length of the screen should be 3 inches greater than its width, what are the dimensions of a 15-inch monitor?

🔗

Solution.

We express the two dimensions of the screen in terms of a single variable:

🔗

\begin{matrix} Width of screen: w \\ Length of screen: w + 3 \end{matrix}

We apply the Pythagorean theorem to write an equation

[TK] :

🔗

w^{2} + (w + 3)^{2} = 15^{2}

To solve this equation, we begin by simplifying the left side.

🔗

\begin{aligned} w^{2} + w^{2} + 6 w + 9 & = 225 & Write the equation in standard form. \\ 2 w^{2} + 6 w - 216 & = 0 & Factor 2 from the left side. \\ 2 (w^{2} + 3 w - 108) & = 0 & Factor the quadratic expression. \\ 2 (w - 9) (w + 12) & = 0 & Set each factor equal to zero. \\ w - 9 = 0 or w + 12 & = 0 & Solve each equation. \\ w = 9 or w & = - 12 \end{aligned}

Because the width of the screen cannot be a negative number, the width is 9 inches, and the length is

w + 3 = 12

inches.

🔗

Checkpoint 3.2.15. Practice 5.

Francine is designing the layout for a botanical garden. The plan includes a square herb garden, with a path 5 feet wide through the center of the garden, as shown at right. To include all the species of herbs, the planted area must be 300 square feet. Find the dimensions of the herb garden.

🔗

Answer.

20 feet by 20 feet

🔗

Subsection 3.2.5 More About Solutions of Quadratic Equations

As we have seen in ehe examples above, the solutions of the quadratic equation

🔗

a (x - r_{1}) (x - r_{2}) = 0

are

r_{1}

and

r_{2} .

This is called the factored form of the quadratic equation. Thus, if we know the two solutions of a quadratic equation, we can work backwards to reconstruct the equation.

🔗

Example 3.2.16.

Find a quadratic equation whose solutions are

\frac{1}{2}

and

- 3 .

🔗

Solution.

Each solution corresponds to a factor of the equation, so the equation must look like this:

🔗

(x - \frac{1}{2}) (x - (- 3)) = 0

or, simplifying:

🔗

(x - \frac{1}{2}) (x + 3) = 0

We multiply the factors together to obtain

🔗

x^{2} + \frac{5}{2} x - \frac{3}{2} = 0

This is an equation that works, but we can make a "nicer" one if we clear the fractions. We can multiply both sides of the equation by 2. We know that multiplying by a constant does not change the solutions of the equation.

🔗

\begin{aligned} 2 (x^{2} + \frac{5}{2} x - \frac{3}{2}) & = 2 (0) \\ 2 x^{2} + 5 x - 3 & = 0 \end{aligned}

By factoring, we can check that this equation really does have the given solutions.

🔗

0 = 2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)

From here, you can see that the solutions are indeed

\frac{1}{2}

and

- 3 .

🔗

Checkpoint 3.2.17. Practice 6.

Find a quadratic equation with integer coefficients whose solutions are

\frac{2}{3}

and

- 5 .

🔗

Answer.

3 x^{2} + 13 x - 10 = 0

🔗

A quadratic equation in one variable always has two solutions. In some cases, the solutions may be equal. For example, the equation

🔗

x^{2} - 2 x + 1 = 0

can be solved by factoring as follows:

🔗

\begin{aligned} (x - 1) (x - 1) & = 0 & Apply the zero-factor principle. \\ x - 1 = 0 or x - 1 & = 0 \end{aligned}

Both of these equations have solution

x = 1 .

We say that 1 is a solution of multiplicity two, meaning that it occurs twice as a solution of the quadratic equation.

🔗

Checkpoint 3.2.18. QuickCheck 4.

True or false.

🔗

We find the $x$ -intercepts of a graph by setting $x = 0 .$
If $z$ is a solution of a quadratic equation, then $(x - z)$ is a factor of the left side in standard form.
We can factor a constant from both sides of a quadratic equation without changing its solutions.
To solve a quadratic equation by factoring, we should factor each side of the equation.

🔗

Exercises 3.2.6 Problem Set 3.2

Warm Up

Exercise Group.

For Problems 1-4, write each product as a polynomial in simplest form.

🔗

1.

$(b + 6) (2 b - 3)$
$(3 z - 8) (4 z - 1)$

🔗

2.

$(4 z - 3)^{2}$
$(2 d + 8)^{2}$

🔗

3.

$3 p (2 p - 5) (p - 3)$
$2 v (v + 4) (3 v - 4)$

🔗

4.

$- 50 (1 + r)^{2}$
$12 (1 - t))^{2}$

🔗

5.

Factor if possible.

🔗

$x^{2} - 16$
$x^{2} - 16 x$
$x^{2} - 8 x + 16$
$x^{2} + 16$

🔗

6.

Solve if possible.

🔗

$x^{2} - 16 = 0$
$x^{2} - 16 x = 0$
$x^{2} - 8 x + 16 = 0$
$x^{2} + 16 = 0$

🔗

Exercise Group.

For Problems 7-12, factor completely.

🔗

7.

x^{2} - 7 x + 10

🔗

8.

x^{2} - 225

🔗

9.

w^{2} - 4 w - 32

🔗

10.

2 z^{2} + 11 z - 40

🔗

11.

9 n^{2} + 24 n + 16

🔗

12.

4 n^{2} - 28 n + 49

🔗

Skills Practice

Exercise Group.

For Problems 13–20, solve by factoring.

🔗

13.

2 a^{2} + 5 a - 3 = 0

🔗

14.

3 b^{2} - 4 b - 4 = 0

🔗

15.

2 x^{2} - 6 x = 0

🔗

16.

3 y^{2} - 6 y = - 3

🔗

17.

x (2 x - 3) = - 1

🔗

18.

t (t - 3) = 2 (t - 3)

🔗

19.

z (3 z + 2) = (z + 2)^{2}

🔗

20.

(v + 2) (v - 5) = 8

🔗

Exercise Group.

For problems 21-24, solve by extracting roots.

🔗

21.

3 (8 x - 7)^{2} = 24

🔗

22.

81 {(x + \frac{1}{3})}^{2} = 1

🔗

23.

(a x - b)^{2} = 25

🔗

24.

100 = π x^{2} - 16 π

🔗

Exercise Group.

For problems 25 and 26, graph the equation and locate the

x

-intercepts of the graph. Use the

x

-intercepts to write the quadratic expression in factored form.

🔗

25.

y = 0.1 (x^{2} - 3 x - 270)

🔗

26.

y = - 0.06 (x^{2} - 22 x - 504)

🔗

27.

Use technology to graph all three equations in the same window. What do you notice about the

x

-intercepts?

🔗

$y = x^{2} + 2 x - 15$
$y = 3 (x^{2} + 2 x - 15)$
$y = 0.2 (x^{2} + 2 x - 15)$

🔗

28.

Write a quadratic equation with the given solutions. Give your answers in standard form with integer coefficients.

🔗

$- 2$ and $1$
$- 3$ and $\frac{1}{2}$
$\frac{- 1}{4}$ and $\frac{3}{2}$

🔗

Applications

29.

Delbert stands at the top of a 300-foot cliff and throws his algebra book directly upward with a velocity of 20 feet per second. The height of his book above the ground

t

seconds later is given by the equation

🔗

h = - 16 t^{2} + 20 t + 300

where

h

is in feet.

🔗

Use your graphing utility to make a table of values for the height equation, with increments of 0.5 second.
Graph the height equation. Use your table of values to help you choose appropriate WINDOW settings.
What is the highest altitude Delbert’s book reaches? When does it reach that height? Use the TRACE feature to find approximate answers first. Then use the Table feature to improve your estimate.
When does Delbert’s book pass him on its way down? (Delbert is standing at a height of 300 feet.) Use the intersect command.
Write and solve an equation to answer the question: How long will it take Delbert’s book to hit the ground at the bottom of the cliff?

🔗

30.

The annual increase

I

in the deer population in a national park is given by the formula

🔗

I = 1.2 x - 0.0002 x^{2}

where

x

is the size of the population that year.

🔗

Make a table of values for $I$ for $0 \leq x \leq 7000.$ Use increments of 500 in $x .$
How much will a population of 2000 deer increase? A population of 5000 deer? A population of 7000 deer?
Use your calculator to graph the annual increase versus the size of the population, $x,$ for $0 \leq x \leq 7000.$ Use your table from part (b) to help you choose appropriate values for Ymin and Ymax.
What do the $x$ -intercepts tell us about the deer population?
Estimate the population size that results in the largest annual increase. What is that increase?

🔗

31.

One end of a ladder is 10 feet from the base of a wall, and the other end reaches a window in the wall. The ladder is 2 feet longer than the height of the window.

🔗

Choose a variable for the height of the window. Make a sketch of the situation described, and label the sides of a right triangle.
Write a quadratic equation about the height of the window.
Solve your equation to find the height of the window.

🔗

32.

Irene would like to enclose two adjacent chicken coops of equal size against the henhouse wall. She has 66 feet of chicken wire fencing, and she would like the total area of the two coops to be 360 square feet. What should the dimensions of the chicken coops be?

🔗

We’ll use three methods to solve this problem: a table of values, a graph, and an algebraic equation.

🔗

Make a table by hand that shows the areas of coops of various widths, as shown below.

Width Length Area

$4$ $27$ $216$

$6$ $24$ $288$

Continue the table until you find a pair of chicken coops whose total area is 360 square feet. (Be careful computing the length of each chicken coop: look at the diagram above.)
Write an expression for the length of each of the two coops if their width is $x .$ Then write an expression for the combined area of the coops if their width is $x .$ Graph the equation for $A,$ and use the graph to find the pair of coops whose combined area is 360 square feet. (Is there more than one solution?)
Write an equation for the area $A$ of the two coops in terms of their width, $x .$ Solve your equation algebraically for $A = 360.$

🔗

33.

A box is made from a square piece of cardboard by cutting 2-inch squares from each corner and then turning up the edges.

🔗

If the piece of cardboard is $x$ inches square, write expressions for the length, width, and height of the box. Then write an expression for the volume, $V,$ of the box in terms of $x .$
Use your calculator to make a table of values showing the volumes of boxes made from cardboard squares of side 4 inches, 5 inches, and so on.
Graph your expression for the volume on your calculator. What is the smallest value of $x$ that makes sense for this problem?
Use your table or your graph to find what size cardboard you need to make a box with volume 50 cubic inches.
Write and solve a quadratic equation to answer part (d).

🔗

34.

A length of rain gutter is made from a piece of aluminum 6 feet long and 1 foot wide.

🔗

If a strip of width $x$ is turned up along each long edge, write expressions for the length, width and height of the gutter. Then write an expression for the volume $V$ of the gutter in terms of $x .$
Use your calculator to make a table of values showing the volumes of various rain gutters formed by turning up edges of 0.1 foot, 0.2 foot, and so on.
Graph your expression for the volume. What happens to $V$ as $x$ increases?
Use your table or your graph to discover how much metal should be turned up along each long edge so that the gutter has a capacity of $\frac{3}{4}$ cubic foot of rainwater.
Write and solve a quadratic equation to answer part (d).

🔗

You have attempted 1 of 1 activities on this page.

🔗

Prev Top Next

Width	Length	Area
$4$	$27$	$216$
$6$	$24$	$288$