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3.2 Quadratic Equations, Functions, Zeros, and Models

Find zeros of quadratic functions and solve quadratic equations by using the principle of zero products, by using the principle of square roots, by completing the square, and by using the quadratic formula.
Solve equations that are reducible to quadratic.
Solve applied problems using quadratic equations.

Quadratic Equations and Quadratic Functions

In this section, we will explore the relationship between the solutions of quadratic equations and the zeros of quadratic functions. We define quadratic equations and quadratic functions as follows.

A quadratic equation written in the form ax2+bx+c=0 $a x^{2} + b x + c = 0$ is said to be in standard form.

The zeros of a quadratic function f(x)=ax2+bx+c $f (x) = a x^{2} + b x + c$ are the solutions of the associated quadratic equation ax2+bx+c=0 $a x^{2} + b x + c = 0$ . (These solutions are sometimes called roots of the equation.) Quadratic functions can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions. If the zeros or solutions are real numbers, they are also the first coordinates of the x-intercepts of the graph of the quadratic function.

The following principles allow us to solve many quadratic equations.

Example 1

Solve: 2x2−x=3 $2 x^{2} - x = 3$ .

Now Try Exercise 3.

Example 2

Solve: $2 x^{2} - 10 = 0$ .

Now Try Exercise 7.

We have seen that some quadratic equations can be solved by factoring and using the principle of zero products. For example, consider the equation $x^{2} - 3 x - 4 = 0$ :

$\begin{matrix} x^{2} - 3 x - 4 & = & 0 \\ (x + 1) (x - 4) & = & 0 & Factoring \\ x + 1 = 0 & or & x - 4 & = & 0 & Using the principle of zero products \\ x = - 1 & or & x & = & 4. \end{matrix}$

The equation $x^{2} - 3 x - 4 = 0$ has two real-number solutions, −1 and 4. These are the zeros of the associated quadratic function $f (x) = x^{2} - 3 x - 4$ and the first coordinates of the x-intercepts of the graph of this function. (See Fig. 1.)

Next, consider the equation $x^{2} - 6 x + 9 = 0$ . Again, we factor and use the principle of zero products:

$\begin{array}{l} \begin{array}{rcll} x^{2} - 6 x + 9 & = & 0 \\ (x - 3) (x - 3) & = & 0 & Factoring \end{array} \\ \begin{array}{rclcrcll} x - 3 & = & 0 & or & x - 3 & = & 0 & Using the principle of zero products \\ x & = & 3 & or & x & = & 3. \end{array} \end{array}$

The equation $x^{2} - 6 x + 9 = 0$ has one real-number solution, 3. It is the zero of the quadratic function $g (x) = x^{2} - 6 x + 9$ and the first coordinate of the x-intercept of the graph of this function. (See Fig. 2.)

The principle of square roots can be used to solve quadratic equations like $x^{2} + 13 = 0$ :

$\begin{array}{rcll} x^{2} + 13 & = & 0 \\ x^{2} & = & - 13 \\ x & = & \pm \sqrt{- 13} & Using the principle of square roots \\ x & = & \pm \sqrt{13} i . & \sqrt{- 13} = \sqrt{- 1} \cdot \sqrt{13} = i \cdot \sqrt{13} = \sqrt{13} i \end{array}$

The equation has two imaginary-number solutions, $- \sqrt{13} i$ and $\sqrt{13} i$ . These are the zeros of the associated quadratic function $h (x) = x^{2} + 13$ . Since the zeros are not real numbers, the graph of the function has no x-intercepts. (See Fig. 3.)

Completing the Square

Neither the principle of zero products nor the principle of square roots would yield the exact zeros of a function like $f (x) = x^{2} - 6 x - 10$ or the exact solutions of the associated equation $x^{2} - 6 x - 10 = 0$ . If we wish to find exact zeros or solutions, we can use a procedure called completing the square and then use the principle of square roots.

Example 3

Find the zeros of $f (x) = x^{2} - 6 x - 10$ by completing the square.

Solution

We find the values of x for which $f (x) = 0;$ that is, we solve the associated equation $x^{2} - 6 x - 10 = 0$ . Our goal is to find an equivalent equation of the form $x^{2} + b x + c = d$ in which $x^{2} + b x + c$ is a perfect square. Since

$x^{2} + b x + {(\frac{b}{2})}^{2} = {(x + \frac{b}{2})}^{2},$

the number c is found by taking half the coefficient of the x-term and squaring it. Thus for the equation $x^{2} - 6 x - 10 = 0$ , we have

$\begin{array}{rcll} x^{2} - 6 x - 10 & = & 0 \\ x^{2} - 6 x & = & 10 & Adding 10 \\ x^{2} - 6 x + 9 & = & 10 + 9 & Adding 9 to complete the square : \\ {(\frac{b}{2})}^{2} = {(\frac{- 6}{2})}^{2} = {(- 3)}^{2} = 9 \\ x^{2} - 6 x + 9 & = & 19. \end{array}$

Because $x^{2} - 6 x + 9$ is a perfect square, we are able to write it as ${(x - 3)}^{2}$ , the square of a binomial. We can then use the principle of square roots to finish the solution:

$\begin{array}{rcll} {(x - 3)}^{2} & = & 19 & Factoring \\ x - 3 & = & \pm \sqrt{19} & Using the principle of square roots \\ x & = & 3 \pm \sqrt{19} . & Adding 3 \end{array}$

Therefore, the solutions of the equation are $3 + \sqrt{19}$ and $3 - \sqrt{19}$ , or simply $3 \pm \sqrt{19}$ . The zeros of $f (x) = x^{2} - 6 x - 10$ are also $3 + \sqrt{19}$ and $3 - \sqrt{19}$ , or $3 \pm \sqrt{19}$ .

We can find decimal approximations for $3 \pm \sqrt{19}$ using a calculator:

$3 + \sqrt{19} \approx 7.359 and 3 - \sqrt{19} \approx - 1.359.$

The zeros are approximately 7.359 and −1.359.

Now Try Exercise 31.

Before we can complete the square, the coefficient of the $x^{2}$ -term must be 1. When it is not, we divide by the $x^{2}$ -coefficient on both sides of the equation.

Example 4

Solve: $2 x^{2} - 1 = 3 x$ .

Solution

We have

$\begin{array}{rcll} 2 x^{2} - 1 & = & 3 x \\ 2 x^{2} - 3 x - 1 & = & 0 & Subtracting 3 x . We are unable to factor the result. \\ 2 x^{2} - 3 x & = & 1 & Adding 1 \\ x^{2} - \frac{3}{2} x & = & \frac{1}{2} & Dividing by 2 to make the x^{2} -coefficient 1 \\ x^{2} - \frac{3}{2} x + \frac{9}{16} & = & \frac{1}{2} + \frac{9}{16} & Completing the square : \frac{1}{2} (- \frac{3}{2}) = - \frac{3}{4} and {(- \frac{3}{4})}^{2} = \frac{9}{16}; adding \frac{9}{16} \\ {(x - \frac{3}{4})}^{2} & = & \frac{17}{16} & Factoring and simplifying \\ x - \frac{3}{4} & = & \pm \frac{\sqrt{17}}{4} & \begin{array}{l} Using the principle of square roots and the quotient rule for radicals \end{array} \\ x & = & \frac{3}{4} \pm \frac{\sqrt{17}}{4} & Adding \frac{3}{4} \\ x & = & \frac{3 \pm \sqrt{17}}{4} . \end{array}$

The solutions are

$\frac{3 + \sqrt{17}}{4} and \frac{3 - \sqrt{17}}{4}, or \frac{3 \pm \sqrt{17}}{4} .$

Now Try Exercise 35.

Using the Quadratic Formula

Because completing the square works for any quadratic equation, it can be used to solve the general quadratic equation $a x^{2} + b x + c = 0$ for x. The result will be a formula that can be used to solve any quadratic equation quickly.

Consider any quadratic equation in standard form:

$a x^{2} + b x + c = 0, a \neq 0.$

For now, we assume that $a > 0$ and solve by completing the square. As the steps are carried out, compare them with those of Example 4.

$\begin{array}{rcll} a x^{2} + b x + c & = & 0 & Standard form \\ a x^{2} + b x & = & - c & Adding - c \\ x^{2} + \frac{b}{a} x & = & - \frac{c}{a} & Dividing by a \end{array}$

Half of $\frac{b}{a}$ is $\frac{b}{2 a}$ , and ${(\frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}}$ . Thus we add $\frac{b^{2}}{4 a^{2}}$ :

$\begin{array}{rcl} x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} & = & - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} & Adding \frac{b^{2}}{4 a^{2}} to complete the square \\ {(x + \frac{b}{2 a})}^{2} & = & - \frac{4 a c}{4 a^{2}} + \frac{b^{2}}{4 a^{2}} & \begin{array}{l} Factoring on the left; finding a \\ common denominator on the \\ right : - \frac{c}{a} = - \frac{c}{a} \cdot \frac{4 a}{4 a} = - \frac{4 a c}{4 a^{2}} \end{array} \\ {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2} - 4 a c}{4 a^{2}} \\ x + \frac{b}{2 a} & = & \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} & \begin{array}{l} Using the principle of square roots \\ and the quotient rule for radicals . \\ Since a > 0, \sqrt{4 a^{2}} = 2 a . \end{array} \\ x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} & Adding - \frac{b}{2 a} \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} . \end{array}$

It can also be shown that this result holds if $a < 0$ .

Example 5

Solve $3 x^{2} + 2 x = 7$ . Find exact solutions and approximate solutions rounded to three decimal places.

Solution

After writing the equation in standard form, we are unable to factor, so we identify a, b, and c in order to use the quadratic formula:

$\begin{array}{rcl} 3 x^{2} + 2 x & = & 7 \\ 3 x^{2} + 2 x - 7 & = & 0; \\ a = 3, b = 2, c & = & - 7. \end{array}$

We then use the quadratic formula:

$\begin{array}{rcl} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- 2 \pm \sqrt{2^{2} - 4 (3) (- 7)}}{2 (3)} & Substituting \\ = & \frac{- 2 \pm \sqrt{4 + 84}}{6} = \frac{- 2 \pm \sqrt{88}}{6} \\ = & \frac{- 2 \pm \sqrt{4 \cdot 22}}{6} = \frac{- 2 \pm 2 \sqrt{22}}{6} = \frac{2 (- 1 \pm \sqrt{22})}{2 \cdot 3} \\ = & \frac{2}{2} \cdot \frac{- 1 \pm \sqrt{22}}{3} = \frac{- 1 \pm \sqrt{22}}{3} . \end{array}$

The exact solutions are

$\frac{- 1 - \sqrt{22}}{3} and \frac{- 1 + \sqrt{22}}{3.}$

Using a calculator, we approximate the solutions to be −1.897 and 1.230.

Now Try Exercise 41.

Example 6

Solve: $x^{2} + 5 x + 8 = 0$ .

Now Try Exercise 47.

The Discriminant

From the quadratic formula, we know that the solutions $x_{1}$ and $x_{2}$ of a quadratic equation are given by

$x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} .$

The expression $b^{2} - 4 a c$ shows the nature of the solutions. This expression is called the discriminant. If it is 0, then it makes no difference whether we choose the plus sign or the minus sign in the formula. That is, $x_{1} = - \frac{b}{2 a} = x_{2}$ , so there is just one solution. In this case, we sometimes say that there is one repeated real solution. If the discriminant is positive, there will be two real solutions. If it is negative, we will be taking the square root of a negative number; hence there will be two imaginary-number solutions, and they will be complex conjugates.

In Example 5, the discriminant, 88, is positive, indicating that there are two different real-number solutions. The negative discriminant, −7, in Example 6 indicates that there are two different imaginary-number solutions.

Equations Reducible to Quadratic

Some equations can be treated as quadratic, provided we make a suitable substitution. For example, consider the following:

The equation $u^{2} - 5 u + 4 = 0$ can be solved for u by factoring or using the quadratic formula. Then we can reverse the substitution, replacing u with $x^{2}$ , and solve for x. Equations like the one above are said to be reducible to quadratic, or quadratic in form.

Example 7

Solve: $x^{4} - 5 x^{2} + 4 = 0$ .

Now Try Exercise 79.

Example 8

Solve: $t^{2 / 3} - 2 t^{1 / 3} - 3 = 0$ .

Solution

We let $u = t^{1 / 3}$ and substitute:

$\begin{array}{l} \begin{array}{rcll} t^{2 / 3} - 2 t^{1 / 3} - 3 & = & 0 \\ {(t^{1 / 3})}^{2} - 2 t^{1 / 3} - 3 & = & 0 \\ u^{2} - 2 u - 3 & = & 0 & Substituting u for t^{1 / 3} \\ (u + 1) (u - 3) & = & 0 & Factoring \end{array} \\ \begin{array}{rclcrcll} u + 1 & = & 0 & or & u - 3 & = & 0 & Using the principle of zero products \\ u & = & - 1 & or & u & = & 3. \end{array} \end{array}$

Now we must solve for the original variable, t. We substitute $t^{1 / 3}$ for u and solve for t:

$\begin{array}{rclcrcll} t^{1 / 3} & = & - 1 & or & t^{1 / 3} & = & 3 \\ {(t^{1 / 3})}^{3} & = & {(- 1)}^{3} & or & {(t^{1 / 3})}^{3} & = & 3^{3} & Cubing on both sides \\ t & = & - 1 & or & t & = & 27. \end{array}$

The solutions are −1 and 27.

Now Try Exercise 87.

Applications

Example 9

Museums in China The number of museums in China increased from approximately 2000 in the year 2000 to over 3500 by the end of 2012. In 2012, a record 451 new museums opened. For comparison, in the United States, only 20–40 new museums were opened per year from 2000 to 2008. The function

$h (x) = 30.992 x^{2} + 4.108 x + 2294.594$

can be used to estimate the number of museums in China, x years after 2005. (Source: The Economist/www.economist.com) Use this function to answer the following.

Estimate the number of museums that will be in China in 2017 if the number of new museums that open per year continues at the same rate.
In what year was the number of museums in China 2600?

Solution

For 2017, $x = 2017 - 2005 = 12$ . We substitute 12 for x and find $h (12)$ :

$\begin{array}{rcl} h (x) & = & 30.992 x^{2} + 4.108 x + 2294.594 \\ h (12) & = & 30.992 {(12)}^{2} + 4.108 (12) + 2294.594 \\ h (12) & = & 4462.848 + 49.296 + 2294.594 \approx 6807. \end{array}$

In 2017, there will be approximately 6807 museums in China.
We substitute 2600 for $h (x)$ and solve for x:

$\begin{array}{rcl} h (x) & = & 30.992 x^{2} + 4.108 x + 2294.594 \\ 2600 & = & 30.992 x^{2} + 4.108 x + 2294.594 \\ 0 & = & 30.992 x^{2} + 4.108 x - 305.406. \end{array}$

We then use the quadratic formula, with $a = 30.992$ , $b = 4.108$ , and $c = - 305.406$ :

$\begin{array}{rcl} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x & = & \frac{- 4.108 \pm \sqrt{{(4.108)}^{2} - 4 (30.992) (- 305.406)}}{2 (30.992)} \\ x & = & \frac{- 4.108 \pm \sqrt{37, 877.44667}}{61.984} \\ x & = & 3.074 or x = - 3.206. \end{array}$

Because we are looking for a year after 2005, we use the positive solution. Thus there were about 2600 museums in China 3 years after 2005, or in 2008.

Now Try Exercise 97.

Example 10

Sales of New Homes Sales of new homes have increased in recent years. The function

$h (x) = 22.1 x^{2} - 72.2 x + 371.9$

can be used to estimate the sales of new homes, in thousands, in the United States, where x is the number of years after 2009 (Source: IHS Global Insight). In what year were the number of sales of new homes about 563,400, or 563.4 thousands?

Solution

We substitute 563.4 for $h (x)$ and solve for x:

$\begin{array}{rcl} 563.4 & = & 22.1 x^{2} - 72.2 x + 371.9 \\ 0 & = & 22.1 x^{2} - 72.2 x - 191.5. \end{array}$

We then use the quadratic formula, with $a = 22.1$ , $b = - 72.2$ , and $c = - 191.5$ :

$\begin{array}{rcl} x & = & \frac{- (- 72.2) \pm \sqrt{{(- 72.2)}^{2} - 4 (22.1) (- 191.5)}}{2 (22.1)} & Substituting \\ x & = & \frac{72.2 \pm \sqrt{22, 141.44}}{44.2} \\ x & = & 5 or x = - 1.7 \end{array}$

Because we are looking for a year after 2009, we use the positive solution. Thus there were about 563,400 sales of new homes 5 years after 2009, or in 2014.

Now Try Exercise 95.

Example 11

Train Speeds Two trains leave a station at the same time. One train travels due west, and the other travels due south. The train traveling west travels 20 km/h faster than the train traveling south. After 2 hr, the trains are 200 km apart. Find the speed of each train.

Solution

Familiarize. First, we make a drawing. We let $r =$ the speed of the train traveling south, in kilometers per hour. Then $r + 20 =$ the speed of the train traveling west, in kilometers per hour. We use the motion formula $d = r t$ , where d is the distance, r is the rate (or speed), and t is the time. After 2 hr, the train traveling south has traveled 2r kilometers, and the train traveling west has traveled $2 (r + 20)$ kilometers. We add these distances to the drawing.
Translate. We use the Pythagorean theorem, $a^{2} + b^{2} = c^{2}$ , where a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse:

JUST IN TIME 25

${[2 (r + 20)]}^{2} + {(2 r)}^{2} = 200^{2} .$
Carry out. We solve the equation:

$\begin{array}{l} \begin{array}{rcll} {[2 (r + 20)]}^{2} + {(2 r)}^{2} & = & 200^{2} \\ 4 (r^{2} + 40 r + 400) + 4 r^{2} & = & 40,000 \\ 4 r^{2} + 160 r + 1600 + 4 r^{2} & = & 40,000 \\ 8 r^{2} + 160 r + 1600 & = & 40,000 & Collecting like terms \\ 8 r^{2} + 160 r - 38,400 & = & 0 & Subtracting 40,000 \\ r^{2} + 20 r - 4800 & = & 0 & Dividing by 8 \\ (r + 80) (r - 60) & = & 0 & Factoring \end{array} \\ \begin{array}{rclcrcll} r + 80 & = & 0 & or & r - 60 & = & 0 & Principle of zero products \\ r & = & - 80 & or & r & = & 60. \end{array} \end{array}$
Check. Since speed cannot be negative, we need check only 60. If the speed of the train traveling south is 60 km/h, then the speed of the train traveling west is $60 + 20$ , or 80 km/h. In 2 hr, the train heading south travels $60 \cdot 2$ , or 120 km, and the train heading west travels $80 \cdot 2$ , or 160 km. Then they are $\sqrt{120^{2} + 160^{2}}$ , or $\sqrt{40,000}$ , or 200 km apart. The answer checks.
State. The speed of the train heading south is 60 km/h, and the speed of the train heading west is 80 km/h.

Now Try Exercise 101.

Connecting the Concepts Zeros, Solutions, and Intercepts

The zeros of a function $y = f (x)$ are also the solutions of the equation $f (x) = 0$ , and the real-number zeros are the first coordinates of the x-intercepts of the graph of the function.

3.2 Exercise Set

Solve.

1. $(2 x - 3) (3 x - 2) = 0$
2. $(2 x + 3) (5 x - 2) = 0$
3. $x^{2} - 8 x - 20 = 0$
4. $x^{2} + 6 x + 8 = 0$
5. $3 x^{2} + x - 2 = 0$
6. $10 x^{2} - 16 x + 6 = 0$
7. $4 x^{2} - 12 = 0$
8. $6 x^{2} = 36$
9. $3 x^{2} = 21$
10. $2 x^{2} - 20 = 0$
11. $5 x^{2} + 10 = 0$
12. $4 x^{2} + 12 = 0$
13. $x^{2} + 16 = 0$
14. $x^{2} + 25 = 0$
15. $2 x^{2} = 6 x$
16. $18 x + 9 x^{2} = 0$
17. $3 y^{3} - 5 y^{2} - 2 y = 0$
18. $3 t^{3} + 2 t = 5 t^{2}$
19. $7 x^{3} + x^{2} - 7 x - 1 = 0$ (Hint: Factor by grouping.)
20. $3 x^{3} + x^{2} - 12 x - 4 = 0$ (Hint: Factor by grouping.)

In Exercises 21–28, use the given graph to find each of the following: (a) the x-intercept(s) and (b) the zero(s) of the function.

Solve by completing the square to obtain exact solutions.

29. $x^{2} + 6 x = 7$
30. $x^{2} + 8 x = - 15$
31. $x^{2} = 8 x - 9$
32. $x^{2} = 22 + 10 x$
33. $x^{2} + 8 x + 25 = 0$
34. $x^{2} + 6 x + 13 = 0$
35. $3 x^{2} + 5 x - 2 = 0$
36. $2 x^{2} - 5 x - 3 = 0$

Use the quadratic formula to find exact solutions.

37. $x^{2} - 2 x = 15$
38. $x^{2} + 4 x = 5$
39. $5 m^{2} + 3 m = 2$
40. $2 y^{2} - 3 y - 2 = 0$
41. $3 x^{2} + 6 = 10 x$
42. $3 t^{2} + 8 t + 3 = 0$
43. $x^{2} + x + 2 = 0$
44. $x^{2} + 1 = x$
45. $5 t^{2} - 8 t = 3$
46. $5 x^{2} + 2 = x$
47. $3 x^{2} + 4 = 5 x$
48. $2 t^{2} - 5 t = 1$
49. $x^{2} - 8 x + 5 = 0$
50. $x^{2} - 6 x + 3 = 0$
51. $3 x^{2} + x = 5$
52. $5 x^{2} + 3 x = 1$
53. $2 x^{2} + 1 = 5 x$
54. $4 x^{2} + 3 = x$
55. $5 x^{2} + 2 x = - 2$
56. $3 x^{2} + 3 x = - 4$

For each of the following, find the discriminant, $b^{2} - 4 a c$ , and then determine whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.

57. $4 x^{2} = 8 x + 5$
58. $4 x^{2} - 12 x + 9 = 0$
59. $x^{2} + 3 x + 4 = 0$
60. $x^{2} - 2 x + 4 = 0$
61. $5 t^{2} - 7 t = 0$
62. $5 t^{2} - 4 t = 11$

Find the zeros of the function. Give exact answers and approximate solutions rounded to three decimal places when possible.

63. $f (x) = x^{2} + 6 x + 5$
64. $f (x) = x^{2} - x - 2$
65. $f (x) = x^{2} - 3 x - 3$
66. $f (x) = 3 x^{2} + 8 x + 2$
67. $f (x) = x^{2} - 5 x + 1$
68. $f (x) = x^{2} - 3 x - 7$
69. $f (x) = x^{2} + 2 x - 5$
70. $f (x) = x^{2} - x - 4$
71. $f (x) = 2 x^{2} - x + 4$
72. $f (x) = 2 x^{2} + 3 x + 2$
73. $f (x) = 3 x^{2} - x - 1$
74. $f (x) = 3 x^{2} + 5 x + 1$
75. $f (x) = 5 x^{2} - 2 x - 1$
76. $f (x) = 4 x^{2} - 4 x - 5$
77. $f (x) = 4 x^{2} + 3 x - 3$
78. $f (x) = x^{2} + 6 x - 3$

Solve.

79. $x^{4} - 3 x^{2} + 2 = 0$
80. $x^{4} + 3 = 4 x^{2}$
81. $x^{4} + 3 x^{2} = 10$
82. $x^{4} - 8 x^{2} = 9$
83. $y^{4} + 4 y^{2} - 5 = 0$
84. $y^{4} - 15 y^{2} - 16 = 0$
85. $x - 3 \sqrt{x} - 4 = 0$

(Hint: Let $u = \sqrt{x}$ .)
86. $2 x - 9 \sqrt{x} + 4 = 0$
87. $m^{2 / 3} - 2 m^{1 / 3} - 8 = 0$

(Hint: Let $u = m^{1 / 3}$ .)
88. $t^{2 / 3} + t^{1 / 3} - 6 = 0$
89. $x^{1 / 2} - 3 x^{1 / 4} + 2 = 0$
90. $x^{1 / 2} - 4 x^{1 / 4} = - 3$
91. ${(2 x - 3)}^{2} - 5 (2 x - 3) + 6 = 0$

(Hint: Let $u = 2 x - 3$ .)
92. ${(3 x + 2)}^{2} + 7 (3 x + 2) - 8 = 0$
93. ${(2 t^{2} + t)}^{2} - 4 (2 t^{2} + t) + 3 = 0$
94. $12 = {(m^{2} - 5 m)}^{2} + (m^{2} - 5 m)$

Funding for Afghan Security. The number of U.S. forces in Afghanistan decreased to approximately 34,000 in 2014 from a high of about 100,000 in 2010. The amount of U.S. funding for Afghan security forces also decreased during this period. The function

$f (x) = - 1.321 x^{2} + 5.156 x + 5.517$

can be used to estimate the amount of U.S. funding for Afghan security forces, in billions of dollars, x years after 2009 (Source: U.S. Department of Defense; Brookings Institution; International Security Assistance Force; ESRI). Use this function for Exercises 95 and 96.

95. In what year was the amount of U.S. funding for Afghan security forces about $10.5 billion?
96. In what year was the amount of U.S. funding for Afghan security forces about $5.0 billion?

Multigenerational Households. After declining between 1940 and 1980, the number of multigenerational American households has been increasing since 1980. The function

$h (x) = 0.012 x^{2} - 0.583 x + 35.727$

can be used to estimate the number of multigenerational households in the United States, in millions, x years after 1940 (Source: Pew Research Center). Use this function for Exercises 97 and 98.

97. In what year were there 40 million multigenerational households?
98. In what year were there 55 million multigenerational households?

Time of a Free Fall. The formula $s = 16 t^{2}$ is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds. Use this formula for Exercises 99 and 100.

99. The Taipei 101 Tower, also known as the Taipei Financial Center, in Taipei, Taiwan, is 1670 ft tall. How long would it take an object dropped from the top to reach the ground?
100. At 630 ft, the Gateway Arch in St. Louis is the tallest man-made monument in the United States. How long would it take an object dropped from the top to reach the ground?
101. The length of a rectangular poster is 1 ft more than the width, and a diagonal of the poster is 5 ft. Find the length and the width.
102. One leg of a right triangle is 7 cm less than the length of the other leg. The length of the hypotenuse is 13 cm. Find the lengths of the legs.
103. One number is 5 greater than another. The product of the numbers is 36. Find the numbers.
104. One number is 6 less than another. The product of the numbers is 72. Find the numbers.
105. Box Construction. An open box is made from a 10-cm by 20-cm piece of tin by cutting a square from each corner and folding up the edges. The area of the resulting base is $96 {cm}^{2}$ . What is the length of the sides of the squares?
106. Petting Zoo Dimensions. At the Glen Island Zoo, 170 m of fencing was used to enclose a rectangular petting area of $1750 m^{2}$ . Find the dimensions of the petting area.
107. Dimensions of a Rug. Find the dimensions of a rectangular Persian rug whose perimeter is 28 ft and whose area is $48 {ft}^{2}$ .
108. Picture Frame Dimensions. The rectangular frame on a picture is 8 in. by 10 in. outside and is of uniform width. What is the width of the frame if $48 {in}^{2}$ of the picture shows?

State whether the function is linear or quadratic.

109. $f (x) = 4 - 5 x$
110. $f (x) = 4 - 5 x^{2}$
111. $f (x) = 7 x^{2}$
112. $f (x) = 23 x + 6$
113. $f (x) = 1.2 x - {(3.6)}^{2}$
114. $f (x) = 2 - x - x^{2}$

Skill Maintenance

Cost of a Super Bowl Ad. The cost of a 30-sec Super Bowl ad has increased more than 70% since 2004. The function

$C (x) = 0.17 x + 2.25$

can be used to estimate the cost of a 30-sec ad, in millions of dollars, x years after 2004 (Source: Katar Media). Use this function for Exercises 115 and 116. [1.2]

115. Estimate the cost of a 30-sec Super Bowl ad in 2014.
116. When will the cost of a 30-sec Super Bowl ad reach $5.0 million?

Determine whether the graph is symmetric with respect to the x-axis, the y-axis, and the origin. [2.4]

117. $3 x^{2} + 4 y^{2} = 5$
118. $y^{3} = 6 x^{2}$

Determine whether the function is even, odd, or neither even nor odd. [2.4]

119. $f (x) = 2 x^{3} - x$
120. $f (x) = 4 x^{2} + 2 x - 3$

Synthesis

For each equation in Exercises 121–124, under the given condition: (a) find k and (b) find a second solution.

121. $k x^{2} - 17 x + 33 = 0;$ one solution is 3
122. $k x^{2} - 2 x + k = 0$ ; one solution is −3
123. $x^{2} - k x + 2 = 0$ ; one solution is $1 + i$
124. $x^{2} - (6 + 3 i) x + k = 0$ ; one solution is 3

Solve.

125. ${(x - 2)}^{3} = x^{3} - 2$
126. ${(x + 1)}^{3} = {(x - 1)}^{3} + 26$
127. $(6 x^{3} + 7 x^{2} - 3 x) (x^{2} - 7) = 0$
128. $(x - \frac{1}{5}) (x^{2} - \frac{1}{4}) + (x - \frac{1}{5}) (x^{2} + \frac{1}{8}) = 0$
129. $x^{2} + x - \sqrt{2} = 0$
130. $x^{2} + \sqrt{5} x - \sqrt{3} = 0$
131. $2 t^{2} + {(t - 4)}^{2} = 5 t (t - 4) + 24$
132. $9 t (t + 2) - 3 t (t - 2) = 2 (t + 4) (t + 6)$
133. $\sqrt{x - 3} - \sqrt[4]{x - 3} = 2$
134. $x^{2} + 3 x + 1 - \sqrt{x^{2} + 3 x + 1} = 8$
135. ${(y + \frac{2}{y})}^{2} + 3 y + \frac{6}{y} = 4$
136. Solve $\frac{1}{2} {at}^{2} + v_{0} t + x_{0} = 0$ for t.

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Table of Contents for
3.2 Quadratic Equations, Functions, Zeros, and Models

3.2 Quadratic Equations, Functions, Zeros, and Models

Quadratic Equations and Quadratic Functions

Example 1

Example 2

Figure 1.

Figure 2.

Figure 3.

Completing the Square

Example 3

Solution

Example 4

Solution

Using the Quadratic Formula

Example 5

Solution

Example 6

The Discriminant

Equations Reducible to Quadratic

Example 7

Example 8

Solution

Applications

Example 9

Solution

Example 10

Solution

Example 11

Solution

3.2 Exercise Set

Skill Maintenance

Synthesis

Table of Contents for 3.2 Quadratic Equations, Functions, Zeros, and Models

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Table of Contents for
3.2 Quadratic Equations, Functions, Zeros, and Models