SECTION 1.5 Quadratic Equations - MR. THOMPSON · Section 1.5 Quadratic Equations 147 Solving Quadratic Equations by the Square Root Property Quadratic equations of the form u2 =

144 Chapter 1 Equations and Inequalities

Quadratic Equations SECTION 1.5

Objectives � Solve quadratic

equations by factoring. � Solve quadratic

equations by the square root property.

� Solve quadratic equations by completing the square.

� Solve quadratic equations using the quadratic formula.

� Use the discriminant to determine the number and type of solutions.

� Determine the most effi cient method to use when solving a quadratic equation.

� Solve problems modeled by quadratic equations.

Equations quadratical? Cheerful news about the square of the hypotenuse? You’ve come to the right place. In this section, we study a number of methods for solving quadratic equations , equations in which the highest exponent on the variable is 2. (Yes, it’s quadratic and not quadratical , despite the latter’s rhyme with mathematical.) We also look at applications of quadratic equations, introducing (cheerfully, of course) the Pythagorean Theorem and the square of the hypotenuse.

We begin by defi ning a quadratic equation.

Defi nition of a Quadratic Equation

A quadratic equation in x is an equation that can be written in the general form

ax2 + bx + c = 0,

where a, b, and c are real numbers, with a � 0. A quadratic equation in x is also called a second-degree polynomial equation in x.

Solving Quadratic Equations by Factoring Here is an example of a quadratic equation in general form:

x2-7x+10=0.

a = 1 b = −7 c = 10

We can factor the left side of this equation. We obtain (x - 5)(x - 2) = 0. If a quadratic equation has zero on one side and a factored expression on the other side, it can be solved using the zero-product principle .

� Solve quadratic equations by factoring.

The Zero-Product Principle

If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.

If AB = 0, then A = 0 or B = 0.

I’m very well acquainted, too, with matters mathematical, I understand equations, both simple and quadratical.About binomial theorem I’m teeming with a lot of news,With many cheerful facts about the square of the hypotenuse.

—Gilbert and Sullivan, The Pirates of Penzance

M04_BLIT2281_06_SE_01-hr.indd 144 13/09/12 4:57 PM

Section 1.5 Quadratic Equations 145

For example, consider the equation (x - 5)(x - 2) = 0. According to the zero-product principle, this product can be zero only if at least one of the factors is zero. We set each individual factor equal to zero and solve each resulting equation for x.

(x - 5)(x - 2) = 0

x - 5 = 0 or x - 2 = 0

x = 5 x = 2

We can check each of these proposed solutions, 5 and 2, in the original quadratic equation, x2 - 7x + 10 = 0. Substitute each one separately for x in the equation.

Check 5: Check 2:

x2 - 7x + 10 = 0 x2 - 7x + 10 = 0

52 - 7 # 5 + 10 � 0 22 - 7 # 2 + 10 � 0

25 - 35 + 10 � 0 4 - 14 + 10 � 0

0 = 0, true 0 = 0, true

The resulting true statements indicate that the solutions are 2 and 5. The solution set is {2, 5}. Note that with a quadratic equation, we can have two solutions, compared to the conditional linear equation that had one.

Solving a Quadratic Equation by Factoring

1. If necessary, rewrite the equation in the general form ax2 + bx + c = 0, moving all nonzero terms to one side, thereby obtaining zero on the other side.

2. Factor completely. 3. Apply the zero-product principle, setting each factor containing a variable

equal to zero. 4. Solve the equations in step 3. 5. Check the solutions in the original equation.

EXAMPLE 1 Solving Quadratic Equations by Factoring

Solve by factoring:

a. 4x2 - 2x = 0 b. 2x2 + 7x = 4.

SOLUTION a. We begin with 4x2 - 2x = 0.

Step 1 Move all nonzero terms to one side and obtain zero on the other side. All nonzero terms are already on the left and zero is on the other side, so we can skip this step.

Step 2 Factor. We factor out 2x from the two terms on the left side.

4x2 - 2x = 0 This is the given equation.

2x(2x - 1) = 0 Factor.

Steps 3 and 4 Set each factor equal to zero and solve the resulting equations.

2x = 0 or 2x - 1 = 0

x = 0 2x = 1

x = 12



The solution set is 50, 126 .

b. Next, we solve 2x2 + 7x = 4.

Step 1 Move all nonzero terms to one side and obtain zero on the other side. Subtract 4 from both sides and write the equation in general form.

2x2 + 7x = 4 This is the given equation. 2x2 + 7x - 4 = 4 - 4 Subtract 4 from both sides. 2x2 + 7x - 4 = 0 Simplify.

Step 2 Factor.

2x2 + 7x - 4 = 0 (2x - 1)(x + 4) = 0

Steps 3 and 4 Set each factor equal to zero and solve the resulting equations.

2x - 1 = 0 or x + 4 = 0 2x = 1 x = -4 x = 1

2

Step 5 Check the solutions in the original equation.

Check 0:

4x2 - 2x = 0 4 # 02 - 2 # 0 � 0 0 - 0 � 0 0 = 0, true

Check 12 :

4x2 - 2x = 0 411

222 - 21122 � 0

41142 - 211

22 � 0

1 - 1 � 0 0 = 0, true

Step 5 Check the solutions in the original equation.

Check 12:

2x2 + 7x = 4 211

222 + 71122 � 4

12 + 72

� 4 4 = 4, true

Check �4:

2x2 + 7x = 4 2(-4)2 + 7(-4) � 4

32 + (-28) � 4 4 = 4, true

The solution set is 5-4, 126 . ● ● ●

Check Point 1 Solve by factoring: a. 3x2 - 9x = 0 b. 2x2 + x = 1.

TECHNOLOGY Graphic Connections

You can use a graphing utility to check the real solutions of a quadratic equation. The real solutions of ax2 � bx � c � 0 correspond to the x@intercepts of the graph of y � ax2 � bx � c. For example, to check the solutions of 2x2 + 7x = 4, or 2x2 + 7x - 4 = 0, graph y = 2x2 + 7x - 4. The U-shaped, bowl-like graph is shown on the right. Note that it is important to have all nonzero terms on one side of the quadratic equation before entering it into the graphing utility. The x@intercepts are -4 and 12 , and the graph of y = 2x2 + 7x - 4 passes through (-4, 0) and 11

2 , 02. This verifi es that 5-4, 126 is the solution set of 2x2 + 7x - 4 = 0, or equivalently, 2x2 + 7x = 4.

x-intercept is .12

x-intercept is −4.

[−5, 2, 1] by [−11, 2, 1]



Solving Quadratic Equations by the Square Root Property Quadratic equations of the form u2 = d, where u is an algebraic expression and d is a nonzero real number, can be solved by the square root property . First, isolate the squared expression u2 on one side of the equation and the number d on the other side. Then take the square root of both sides. Remember, there are two numbers whose square is d. One number is 2d and one is -2d .

We can use factoring to verify that u2 = d has these two solutions.

u2 = d This is the given equation.

u2 - d = 0 Move all terms to one side and obtain zero onthe other side.

1u + 2d21u - 2d2 = 0 Factor. u + 2d = 0 or u - 2d = 0 Set each factor equal to zero.

u = -2d u = 2d Solve the resulting equations.

Because the solutions differ only in sign, we can write them in abbreviated notation as u = {2d. We read this as “ u equals positive or negative the square root of d ” or “ u equals plus or minus the square root of d. ”

Now that we have verifi ed these solutions, we can solve u2 = d directly by taking square roots. This process is called the square root property .

� Solve quadratic equations by the square root property.

The Square Root Property

If u is an algebraic expression and d is a nonzero number, then u2 = d has exactly two solutions:

If u2 = d, then u = 2d or u = -2d .

Equivalently,

If u2 = d, then u = {2d .

Before you can apply the square root property, a squared expression must be isolated on one side of the equation.

EXAMPLE 2 Solving Quadratic Equationsby the Square Root Property

Solve by the square root property:

a. 3x2 - 15 = 0 b. 9x2 + 25 = 0 c. (x - 2)2 = 6.

SOLUTION To apply the square root property, we need a squared expression by itself on one side of the equation.

3x2-15=0

We want x2

by itself.

9x2+25=0

We want x2

by itself.

(x-2)2=6

The squared expressionis by itself.

a. 3x2 - 15 = 0 This is the original equation.

3x2 = 15 Add 15 to both sides.

x2 = 5 Divide both sides by 3.

x = 25 or x = -25 Apply the square root property. Equivalently, x = {15 .

By checking both proposed solutions in the original equation, we can confi rm that the solution set is 5-15,156 or 5{156 .



b. 9x2 + 25 = 0 This is the original equation.

9x2 = -25 Subtract 25 from both sides.

x2 = - 259

Divide both sides by 9.

x = {A- 259

Apply the square root property.

x = { iA259

= { 53

i Express solutions in terms of i.

Because the equation has an x2@term and no x@term, we can check both

proposed solutions, { 53

i, at once.

Check 53

i and � 53

i:

9x2 + 25 = 0

9a{ 53

ib2

+ 25 � 0

9a 259

i2b + 25 � 0

25(–1)+25 � 0

0=0

25i2+25 � 0

i 2 = −1

true

The solutions are - 53

i and 53

i. The solution set is e- 53

i, 53

i f or e{ 53

i f .

c. (x - 2)2 = 6 This is the original equation.

x - 2 = {26 Apply the square root property.

x = 2 { 26 Add 2 to both sides.

By checking both values in the original equation, we can confi rm that the solution set is 52 + 26, 2 - 266 or 52 { 266 . ● ● ●

Check Point 2 Solve by the square root property: a. 3x2 - 21 = 0 b. 5x2 + 45 = 0 c. (x + 5)2 = 11.

Completing the Square How do we solve an equation in the form ax2 + bx + c = 0 if the trinomial ax2 + bx + c cannot be factored? We cannot use the zero-product principle in such a case. However, we can convert the equation into an equivalent equation that can be solved using the square root property. This is accomplished by completing the square .

� Solve quadratic equations by completing the square.


The graph of

y = 9x2 + 25

has no x@intercepts. This shows that

9x2 + 25 = 0

has no real solutions. Example 2(b) algebraically establishes that the solutions are imaginary numbers.

[−3, 3, 1] by [−5, 100, 5]

Completing the Square

If x2 + bx is a binomial, then by adding ab2b2

, which is the square of half the

coeffi cient of x, a perfect square trinomial will result. That is,

x2 + bx + ab2b2

= ax +b2b2

.



Squares and rectangles make it possible to visualize completing the square, as shown in Figure 1.19 .

EXAMPLE 3 Completing the Square

What term should be added to each binomial so that it becomes a perfect square trinomial? Write and factor the trinomial.

a. x2 + 8x b. x2 - 7x c. x2 +35

x

SOLUTION To complete the square, we must add a term to each binomial. The term that should be added is the square of half the coeffi cient of x.

x2+8x

Add A B2 = 42.Add 16 to complete

the square.

82

x2+ x

Add A � B2 = A B2.Add to complete

the square.

310

9100

12

35

x2-7x

Add A B2, or ,to completethe square.

−72

494

3

5

a. The coeffi cient of the x@term in x2 + 8x is 8. Half of 8 is 4, and 42 = 16. Add 16. The result is a perfect square trinomial.

x2 + 8x + 16 = (x + 4)2

b. The coeffi cient of the x@term in x2 - 7x is -7. Half of -7 is - 72

, and

a- 72b2

=494

. Add 494

. The result is a perfect square trinomial.

x2 - 7x +494

= ax -72b2

c. The coeffi cient of the x@term in x2 +35

x is 35

. Half of 35

is 12# 35

, or 310

, and

a 310b2

=9

100. Add

9100

. The result is a perfect square trinomial.

x2 +35

x +9

100= ax +

310b2

● ● ●

xb2

x

b2

Adding this red square

the larger square.

Q Rof area b2

2

Area of blue region:

Q Rx2 + 2 x = x2 + bxb2

completes

FIGURE 1.19 Visualizing completing the square

GREAT QUESTION! I’m not accustomed to factoring perfect square trinomials in which fractions are involved. Is there a rule or an observation that can make the factoring easier?

Yes. The constant in the factorization is always half the coeffi cient of x.

72

x2-7x+ =ax- b2494

72

35

310

Half the coefficient of x, −7, is − . 35

310Half the coefficient of x, , is .

x2+ =ax+ b2x+9

100

Check Point 3 What term should be added to each binomial so that it becomes a perfect square trinomial? Write and factor the trinomial.

a. x2 + 6x b. x2 - 5x c. x2 +23

x



We can solve any quadratic equation by completing the square. If the coeffi cient of the x2@term is one, we add the square of half the coeffi cient of x to both sides of the equation. When you add a constant term to one side of the equation to complete the square, be certain to add the same constant to the other side of the equation. These ideas are illustrated in Example 4.

EXAMPLE 4 Solving a Quadratic Equation by Completing the Square

Solve by completing the square: x2 - 6x + 4 = 0.

SOLUTION We begin by subtracting 4 from both sides. This is done to isolate the binomial x2 - 6x, so that we can complete the square.

x2 - 6x + 4 = 0 This is the original equation. x2 - 6x = -4 Subtract 4 from both sides.

Next, we work with x2 - 6x = -4 and complete the square. Find half the coeffi cient of the x@term and square it. The coeffi cient of the x@term is -6. Half of -6 is -3 and (-3)2 = 9. Thus, we add 9 to both sides of the equation.

x2 - 6x + 9 = -4 + 9 Add 9 to both sides to complete the square.

(x - 3)2 = 5 Factor and simplify.

x - 3 = 25 or x - 3 = -25 Apply the square root property.

x = 3 + 25 x = 3 - 25 Add 3 to both sides in each equation.

The solutions are 3 { 25 and the solution set is 53 + 25, 3 - 256 , or 53 { 256 . ● ● ●

Check Point 4 Solve by completing the square: x2 + 4x - 1 = 0.

If the coeffi cient of the x2@term in a quadratic equation is not 1, you must divide each side of the equation by this coeffi cient before completing the square. For example, to solve 9x2 - 6x - 4 = 0 by completing the square, fi rst divide every term by 9:

9x2

9-

6x9

-49=

09

x2 -69

x -49= 0

x2 -23

x -49= 0.

Now that the coeffi cient of the x2@term is 1, we can solve by completing the square.

EXAMPLE 5 Solving a Quadratic Equation by Completing the Square

Solve by completing the square: 9x2 - 6x - 4 = 0.

SOLUTION

9x2 - 6x - 4 = 0 This is the original equation.

x2 -23

x -49= 0 Divide both sides by 9.

x2 -23

x =49

Add 49 to both sides to isolate the binomial.

x2 -23

x +19=

49

+19

Complete the square: Half of - 23 is - 2

6 , or - 13 , and 1- 1322 = 1

9 .

ax -13b2

=59

Factor and simplify.

GREAT QUESTION! When I solve a quadratic equation by completing the square, doesn’t this result in a new equation? How do I know that the solutions of this new equation are the same as those of the given equation?

When you complete the square for the binomial expression x2 + bx, you obtain a different polynomial. When you solve a quadratic equation by completing the square, you obtain an equation with the same solution set because the constant needed to complete the square is added to both sides .



The solutions are 1 { 25

3 and the solution set is b 1 { 25

3r . ● ● ●

Check Point 5 Solve by completing the square: 2x2 + 3x - 4 = 0.

Solving Quadratic Equations Using the Quadratic Formula We can use the method of completing the square to derive a formula that can be used to solve all quadratic equations. The derivation given below also shows a particular quadratic equation, 3x2 - 2x - 4 = 0, to specifi cally illustrate each of the steps.

x -13= A

59 or x -

13= -A

59

Apply the square root property.

x -13=253 x -

13= -

253

A59

=1519

=153

x =13

+253 x =

13

-253

Add 13 to both sides and solve for x.

x =1 + 25

3 x =

1 - 253

Express solutions with a common denominator.


Obtain a decimal approximation for each solution of

9x2 - 6x - 4 = 0,

the equation in Example 5.

1 + 25

3� 1.1;

1 - 253

� -0.4

y = 9x2 − 6x − 4

x-intercept≈ − 0.4

x-intercept≈ 1.1

[−2, 2, 1] by [−10, 10, 1]

The x@intercepts of y = 9x2 - 6x - 4 verify the solutions.

Deriving the Quadratic Formula

General Form of a Quadratic Equation Comment A Specifi c Example

ax2 + bx + c = 0, a 7 0 This is the given equation. 3x2 - 2x - 4 = 0

x2 +ba

x +ca= 0 Divide both sides by the coeffi cient of x2. x2 -

23

x -43= 0

x2 +ba

x = - ca

Isolate the binomial by adding - ca

on both sides of the equation.

x2 -23

x =43

x2+ +ax+a b2

=– b2b a

c a

b 2a

b 2a

(half)2

Complete the square. Add the square of half the coeffi cient of x to both sides.

x+a– b2

= b2

+a–(half)2

x2-23

13

43

13

x2 +ba

x +b2

4a2 = - ca

+b2

4a2x2 -

23

x +19=

43

+19

ax +b2ab2

= - ca# 4a4a

+b2

4a2

Factor on the left side and obtain a common denominator on the right side. ax -

13b2

=43# 33

+19

ax +b2ab2

=-4ac + b2

4a2

ax +b2ab2

=b2 - 4ac

4a2

Add fractions on the right side. ax -13b2

=12 + 1

9

ax -13b2

=139

x +b2a

= {Bb2 - 4ac

4a2Apply the square root property. x -

13= {A

139

x +b2a

= { 2b2 - 4ac

2a

Take the square root of the quotient, simplifying the denominator. x -

13= {

2133

x =-b2a

{2b2 - 4ac

2aSolve for x by subtracting

b2a

from both sides.

x =13{213

3

x =-b { 2b2 - 4ac

2aCombine fractions on the right side. x =

1 { 2133



The formula shown at the bottom of the left column on the previous page is called the quadratic formula . A similar proof shows that the same formula can be used to solve quadratic equations if a, the coeffi cient of the x2@term, is negative.

� Solve quadratic equations using the quadratic formula.

The Quadratic Formula

The solutions of a quadratic equation in general form ax2 + bx + c = 0, with a � 0, are given by the quadratic formula :

x equals negative b plus or minusthe square root of b2 − 4ac, all

divided by 2a.

–b_2b2-4ac2a

x= .

To use the quadratic formula, rewrite the quadratic equation in general form if necessary. Then determine the numerical values for a (the coeffi cient of the x2@term), b (the coeffi cient of the x@term ), and c (the constant term). Substitute the values of a, b, and c into the quadratic formula and evaluate the expression. The { sign indicates that there are two (not necessarily distinct) solutions of the equation.

EXAMPLE 6 Solving a Quadratic Equation Usingthe Quadratic Formula

Solve using the quadratic formula: 2x2 - 6x + 1 = 0.

SOLUTION The given equation is in general form. Begin by identifying the values for a, b, and c.

2x2-6x+1=0

a = 2 b = −6 c = 1

Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.

x =-b { 2b2 - 4ac

2a Use the quadratic formula.

=-(-6) { 2(-6)2 - 4(2)(1)

2 # 2 Substitute the values for a, b, and c : a = 2, b = -6, and c = 1.

=6 { 236 - 8

4

-(-6) = 6, (-6)2 = (-6)(-6) = 36, and 4(2)(1) = 8.

=6 { 228

4 Complete the subtraction under the radical.

=6 { 227

4 228 = 24 # 7 = 24 27 = 227

=213 { 272

4 Factor out 2 from the numerator.

=3 { 27

2 Divide the numerator and denominator by 2.

The solution set is b 3 + 272

, 3 - 27

2r or b 3 { 27

2r . ● ● ●

TECHNOLOGY You can use a graphing utility to verify that the solutions of 2x2 - 6x + 1 = 0 are

3 { 27

2. Begin by entering

y1 = 2x2 - 6x + 1 in the � Y= � screen. Then evaluate this equation at each of the proposed solutions.

In each case, the value is 0,verifying that the solutionssatisfy 2x2 − 6x + 1 = 0.



Can all irrational solutions of quadratic equations be simplifi ed? No. The following solutions cannot be simplifi ed.

5_2272

–4_3272

Other than 1, terms ineach numerator have no

common factor.

Check Point 6 Solve using the quadratic formula:

2x2 + 2x - 1 = 0.

EXAMPLE 7 Solving a Quadratic Equation Usingthe Quadratic Formula

Solve using the quadratic formula: 3x2 - 2x + 4 = 0.

SOLUTION The given equation is in general form. Begin by identifying the values for a, b, and c.

3x2-2x+4=0

a = 3 b = −2 c = 4

x =-b { 2b2 - 4ac

2a Use the quadratic formula.

=-(-2) { 2(-2)2 - 4(3)(4)

2(3)

Substitute the values for a, b, and c: a = 3, b = -2, and c = 4.

=2 { 24 - 48

6 -(-2) = 2 and (-2)2 = (-2)(-2) = 4.

=2 { 2-44

6

Subtract under the radical. Because the number under the radical sign is negative, the solutions will not be real numbers.

GREAT QUESTION! The simplifi cation of the irrational solutions in Example 6 is kind of tricky. Any suggestions to guide the process?

Many students use the quadratic formula correctly until the last step, where they make an error in simplifying the solutions. Be sure to factor the numerator before dividing the numerator and denominator by the greatest common factor:

=

1

2

6_2274

3_272

2A3_27B4

= =2A3_27B

4Factorfirst Then divide

by the GCF.

.

You cannot divide just one term in the numerator and the denominator by their greatest common factor.

Incorrect!

6 { 227

4=

63{ 227

42

=3 { 227

2

6 { 2274

=6 { 2

12742

=6 { 27

2

GREAT QUESTION! Should I check irrational and imaginary solutions by substitution in the given quadratic equation?

No. Checking irrational and imaginary solutions can be time-consuming. The solutions given by the quadratic formula are always correct, unless you have made a careless error. Checking for computational errors or errors in simplifi cation is suffi cient. =

2 { 2i2116

2�44 = 24(11)(� 1) = 2i211

=211 { i2112

6 Factor 2 from the numerator.

=1 { i211

3 Divide the numerator and denominator by 2.

=13{ i 211

3 Write the complex numbers in standard form.



The solutions are complex conjugates, and the solution set is

b 13

+ i 211

3,

13

- i 211

3r or b 1

3{ i 211

3r . ● ● ●

If ax2 + bx + c = 0 has imaginary solutions, the graph of y = ax2 + bx + c will not have x@intercepts. This is illustrated by the imaginary solutions of 3x2 - 2x + 4 = 0 in Example 7 and the graph of y = 3x2 - 2x + 4 in Figure 1.20 .

Check Point 7 Solve using the quadratic formula:

x2 - 2x + 2 = 0.

The Discriminant The quantity b2 - 4ac, which appears under the radical sign in the quadratic formula, is called the discriminant . Table 1.2 shows how the discriminant of the quadratic equation ax2 + bx + c = 0 determines the number and type of solutions.

y = 3x2 − 2x + 4

[−2, 2, 1] by [−1, 10, 1]

FIGURE 1.20 This graph has no x@intercepts .

� Use the discriminant to determine the number and type of solutions.

Table 1.2 The Discriminant and the Kinds of Solutions to ax2 � bx � c � 0

Discriminant b2 � 4ac

Kinds of Solutions to ax2 � bx � c � 0

Graph of y � ax2 � bx � c

b2 - 4ac 7 0 Two unequal real solutions: If a, b, and c are rational numbers and the discriminant is a perfect square, the solutions are rational. If the discriminant is not a perfect square, the solutions are irrational conjugates.

x

y

Two x-intercepts

b2 - 4ac = 0 One solution (a repeated solution) that is a real number:If a, b, and c are rational numbers, therepeated solution is also a rationalnumber.

x

y

One x-intercept

b2 - 4ac 6 0 No real solution; two imaginary solutions:The solutions are complex conjugates.

x

y

No x-intercepts

EXAMPLE 8 Using the Discriminant

For each equation, compute the discriminant. Then determine the number and type of solutions:

a. 3x2 + 4x - 5 = 0 b. 9x2 - 6x + 1 = 0 c. 3x2 - 8x + 7 = 0.

SOLUTION Begin by identifying the values for a, b, and c in each equation. Then compute b2 - 4ac, the discriminant.

a. 3x2+4x-5=0

a = 3 b = 4 c = −5

GREAT QUESTION! Is the square root sign part of the discriminant?

No. The discriminant is b2 - 4ac. It is not 2b2 - 4ac , so do not give the discriminant as a radical.



Substitute and compute the discriminant:

b2 - 4ac = 42 - 4 # 3(-5) = 16 - (-60) = 16 + 60 = 76.

The discriminant, 76, is a positive number that is not a perfect square. Thus, there are two irrational solutions. (These solutions are conjugates of each other.)

b. 9x2-6x+1=0

a = 9 b = −6 c = 1

Substitute and compute the discriminant:

b2 - 4ac = (-6)2 - 4 # 9 # 1 = 36 - 36 = 0.

The discriminant, 0, shows that there is only one real solution. This real solution is a rational number.

c. 3x2-8x+7=0

a = 3 c = 7b = −8

b2 - 4ac = (-8)2 - 4 # 3 # 7 = 64 - 84 = -20

The negative discriminant, -20, shows that there are two imaginary solutions. (These solutions are complex conjugates of each other.) ● ● ●

Check Point 8 For each equation, compute the discriminant. Then determine the number and type of solutions:

a. x2 + 6x + 9 = 0 b. 2x2 - 7x - 4 = 0 c. 3x2 - 2x + 4 = 0.

Determining Which Method to Use All quadratic equations can be solved by the quadratic formula. However, if an equation is in the form u2 = d, such as x2 = 5 or (2x + 3)2 = 8, it is faster to use the square root property, taking the square root of both sides. If the equation is not in the form u2 = d, write the quadratic equation in general form (ax2 + bx + c = 0). Try to solve the equation by factoring. If ax2 + bx + c cannot be factored, then solve the quadratic equation by the quadratic formula.

Because we used the method of completing the square to derive the quadratic formula, we no longer need it for solving quadratic equations. However, we will use completing the square later in the book to help graph circles and other kinds of equations.

� Determine the most effi cient method to use when solving a quadratic equation.

GREAT QUESTION! Is factoring the most important technique to use when solving a quadratic equation?

No. Even though you fi rst learned to solve quadratic equations by factoring, most quadratic equations in general form cannot be factored. Be sure to apply the quadratic formula when this occurs.



Table 1.3 summarizes our observations about which technique to use when solving a quadratic equation.

Table 1.3 Determining the Most Effi cient Technique to Use when Solving a Quadratic Equation

Description and Form of the Quadratic Equation

Most Effi cientSolution Method

Example

ax2 + bx + c = 0 and ax2 + bx + c can be factored easily.

Factor and use the zero-product principle.

3x2 + 5x - 2 = 0

(3x - 1)(x + 2) = 0

3x - 1 = 0 or x + 2 = 0

x =13 x = -2

ax2 + bx = 0 The quadratic equation has no constant term. (c = 0)

Factor and use the zero-product principle.

6x2 + 9x = 0

3x(2x + 3) = 0

3x = 0 or 2x + 3 = 0

x = 0 2x = -3

x = - 32

ax2 + c = 0 The quadratic equation has no x@term. (b = 0)

Solve for x2 and apply the square root property.

7x2 - 4 = 0

7x2 = 4

x2 =47

x = {A47

= { 2

27= {

2

27# 27

27= {

2277

u2 = d; u is a fi rst-degree polynomial. Use the square root property. (x + 4)2 = 5

x + 4 = {25

x = -4 { 25

ax2 + bx + c = 0 and ax2 + bx + c cannot be factored or the factoring is too diffi cult.

Use the quadratic formula:

x =-b { 2b2 - 4ac

2a.

x2-2x-6=0

a = 1 b = −2 c = −6

x =-(-2) { 2(-2)2 - 4(1)(-6)

2(1)

=2 { 24 + 24

2(1)

=2 { 228

2=

2 { 24272

=2 { 227

2=

211 { 2722

= 1 { 27



Applications

EXAMPLE 9 Blood Pressure and Age

The graphs in Figure 1.21 illustrate that a person’s normal systolic blood pressure, measured in millimeters of mercury (mm Hg), depends on his or her age. (The symbol on the vertical axis shows there is a break in values between 0 and 100. Thus, the fi rst tick mark on the vertical axis represents a blood pressure of 100 mm Hg.) The formula

P = 0.006A2 - 0.02A + 120

models a man’s normal systolic pressure, P, at age A.

a. Find the age, to the nearest year, of a man whose normal systolic blood pressure is 125 mm Hg.

b. Use the graphs in Figure 1.21 to describe the differences between the normal systolic blood pressures of men and women as they age.

SOLUTION a. We are interested in the age of a man with a normal systolic blood pressure

of 125 millimeters of mercury. Thus, we substitute 125 for P in the given formula for men. Then we solve for A, the man’s age.

0=0.006A2-0.02A-5

125=0.006A2-0.02A+120

P=0.006A2-0.02A+120

a = 0.006 b = −0.02 c = −5

� Solve problems modeled by quadratic equations.

160

150

140

130

120

110

50 60 70

Nor

mal

Blo

od P

ress

ure

(mm

Hg)

Age

Normal SystolicBlood Pressure and Age

10 20 30 40 80

100

P

A

Men

Women

FIGURE 1.21

This is the given formula for men.

Substitute 125 for P.

Subtract 125 from both sides and write the quadratic equation in general form.

Because the trinomial on the right side of the equation is prime, we solve using the quadratic formula.

Notice that thevariable is A,rather than the

usual x.A=

–b_2b2-4ac2a

=–(–0.02)_2(–0.02)2-4(0.006)(–5)

2(0.006)

Use the quadratic formula. Substitute the values for a, b, and c: a = 0.006, b = �0.02, and c = �5.

=0.02 { 20.1204

0.012 Use a calculator to

simplify the radicand.

�0.02 { 0.347

0.012 Use a calculator:

20.1204 ? 0.347.

A �0.02 + 0.347

0.012 or A �

0.02 - 0.3470.012

A � 31

Reject this solution.

Age cannot be negative.

A≠ –27

Use a calculator and round to the nearest integer.

TECHNOLOGY On most calculators, here is how to approximate

0.02 + 20.1204

0.012.

If your calculator displaysan open parenthesis after � ,

you’ll need to enter anotherclosed parenthesis here.

Many Graphing Calculators

, .012 ENTER

( .02 + 1 .1204 )

Many Scientific Calculators

, .012 =

( .02 + .1204 1 )



The positive solution, A � 31, indicates that 31 is the approximate age of a man whose normal systolic blood pressure is 125 mm Hg. This is illustrated by the black lines with the arrows on the red graph representing men in Figure 1.22 .

b. Take a second look at the graphs in Figure 1.22 . Before approximately age 50, the blue graph representing women’s normal systolic blood pressure lies below the red graph representing men’s normal systolic blood pressure. Thus, up to age 50, women’s normal systolic blood pressure is lower than men’s, although it is increasing at a faster rate. After age 50, women’s normal systolic blood pressure is higher than men’s. ● ● ●

Check Point 9 The formula P = 0.01A2 + 0.05A + 107 models a woman’s normal systolic blood pressure, P, at age A. Use this formula to fi nd the age, to the nearest year, of a woman whose normal systolic blood pressure is 115 mm Hg. Use the blue graph in Figure 1.22 to verify your solution.

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle , a triangle with one angle measuring 90�. The side opposite the 90� angle is called the hypotenuse . The other sides are called legs . Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure 1.23 .

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem .

160

150

140

130

120

110

50 60 70

Nor

mal

Blo

od P

ress

ure

(mm

Hg)

Age

Normal SystolicBlood Pressure and Age

10 20 30 40 80

100

Men

Age ~ 31~

Bloodpressure:

125

Women

A

P

FIGURE 1.22

Area:16 squareunits

Area:25squareunits

Area:9 squareunits

53

4

FIGURE 1.23 The area of the large square equals the sum of the areas of the smaller squares.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2 + b2 = c2.

Hypotenusec

B

CA bLeg

aLeg

EXAMPLE 10 Using the Pythagorean Theorem a. A wheelchair ramp with a length of 122 inches has a horizontal distance of

120 inches. What is the ramp’s vertical distance? b. Construction laws are very specifi c when it comes to access ramps for people

with disabilities. Every vertical rise of 1 inch requires a horizontal run of 12 inches. Does this ramp satisfy the requirement?

SOLUTION a. Figure 1.24 shows the right triangle that is formed by the ramp, the wall,

and the ground. We can fi nd x, the ramp’s vertical distance, using the Pythagorean Theorem.

120 in.

x 122 in.

FIGURE 1.24



x2 + 1202 = 1222

plus(leg)2 (leg)2 (hypotenuse)2equals

We solve this equation using the square root property.

x2 + 1202 = 1222 This is the equation resulting from the Pythagorean Theorem.

x2 + 14,400 = 14,884 Square 120 and 122. x2 = 484 Isolate x2 by subtracting 14,400 from both sides. x = 2484 or x = -2484 Apply the square root property. x = 22 x = -22

Because x represents the ramp’s vertical distance, we reject the negative value. Thus, the ramp’s vertical distance is 22 inches.

b. Every vertical rise of 1 inch requires a horizontal run of 12 inches. Because the ramp has a vertical distance of 22 inches, it requires a horizontal distance of 22(12) inches, or 264 inches. The horizontal distance is only 120 inches, so this ramp does not satisfy construction laws for access ramps for people with disabilities. ● ● ●

Check Point 10 In a 25-inch television set, the length of the screen’s diagonal is 25 inches. If the screen’s height is 15 inches, what is its width?

A golden rectangle can be a rectangle of any size, but its long side must be � times as long as its short side, where � � 1.6. Artists often use golden rectangles in their work because they are considered to be more visually pleasing than other rectangles.

If a golden rectangle is divided into a square and a rectangle, as in Figure 1.25(a) , the smaller rectangle is a golden rectangle. If the smaller golden rectangle is divided again, the same is true of the yet smaller rectangle, and so on. The process of repeatedly dividing each golden rectangle in this manner is illustrated in Figure 1.25(b) . We’ve also created a spiral by connecting the opposite corners of all the squares with a smooth curve. This spiral matches the spiral shape of the chambered nautilus shell shown in Figure 1.25(c) . The shell spirals out at an ever-increasing rate that is governed by this geometry.

Blitzer Bonus ❘ ❘ Art, Nature, and Quadratic Equations

In The Bathers at Asnières, by the French impressionist Georges Seurat (1859–1891), the artist positions parts of the painting as though they were inside golden rectangles. Bathers at Asnières (1884), Georges Seurat, Art Resource.

Golden Rectangle A

SquareGolden

RectangleB

FIGURE 1.25(a) FIGURE 1.25(b) FIGURE 1.25(c)

In the Exercise Set that follows, you will use the golden rectangles in Figure 1.25(a) to obtain an exact value for �, the ratio of the long side to the short side in a golden rectangle of any size. Your model will involve a quadratic equation that can be solved by the quadratic formula. (See Exercise 139.)



Fill in each blank so that the resulting statement is true.

CONCEPT AND VOCABULARY CHECK

1. An equation that can be written in the general form ax2 + bx + c = 0, a � 0, is called a/an equation.

2. The zero-product principle states that if AB = 0,then .

3. The solutions of ax2 + bx + c = 0 correspond to the for the graph of y = ax2 + bx + c.

4. The square root property states that if u2 = d, then u = .

5. If x2 = 7, then x = . 6. To complete the square on x2 - 3x, add .

7. To complete the square on x2 -45

x, add .

8. To solve x2 + 6x = 7 by completing the square, add to both sides of the equation.

9. To solve x2 -23

x =49

by completing the square, add

to both sides of the equation. 10. The solutions of a quadratic equation in the general

form ax2 + bx + c = 0, a � 0, are given by the

quadratic formula x = .

11. In order to solve 2x2 + 9x - 5 = 0 by the quadratic formula, we use a = , b = , andc = .

12. In order to solve x2 = 4x + 1 by the quadratic formula, we use a = , b = , and c = .

13. x =-(-4) { 2(-4)2 - 4(1)(2)

2(1) simplifi es to

x = .

14. x =-4 { 242 - 4 # 2 # 5

2 # 2 simplifi es to x = .

15. The discriminant of ax2 + bx + c = 0 is defi ned by .

16. If the discriminant of ax2 + bx + c = 0 is negative, the quadratic equation has real solutions.

17. If the discriminant of ax2 + bx + c = 0 is positive, the quadratic equation has real solutions.

18. The most effi cient technique for solving (2x + 7)2 = 25is by using .

19. The most effi cient technique for solving x2 + 5x - 10 = 0 is by using .

20. The most effi cient technique for solving x2 + 8x + 15 = 0is by using .

21. A triangle with one angle measuring 90� is called a/an triangle. The side opposite the 90� angle is called

the . The other sides are called . 22. The Pythagorean Theorem states that in any

triangle, the sum of the squares of the lengths of the equals .

EXERCISE SET 1.5

Practice Exercises Solve each equation in Exercises 1–14 by factoring.

1. x2 - 3x - 10 = 0 2. x2 - 13x + 36 = 0 3. x2 = 8x - 15 4. x2 = -11x - 10 5. 6x2 + 11x - 10 = 0 6. 9x2 + 9x + 2 = 0 7. 3x2 - 2x = 8 8. 4x2 - 13x = -3 9. 3x2 + 12x = 0 10. 5x2 - 20x = 0 11. 2x(x - 3) = 5x2 - 7x 12. 16x(x - 2) = 8x - 25 13. 7 - 7x = (3x + 2)(x - 1) 14. 10x - 1 = (2x + 1)2

Solve each equation in Exercises 15–34 by the square root property.

15. 3x2 = 27 16. 5x2 = 45 17. 5x2 + 1 = 51 18. 3x2 - 1 = 47 19. 2x2 - 5 = -55 20. 2x2 - 7 = -15 21. (x + 2)2 = 25 22. (x - 3)2 = 36 23. 3(x - 4)2 = 15 24. 3(x + 4)2 = 21 25. (x + 3)2 = -16 26. (x - 1)2 = -9 27. (x - 3)2 = -5 28. (x + 2)2 = -7 29. (3x + 2)2 = 9 30. (4x - 1)2 = 16 31. (5x - 1)2 = 7 32. (8x - 3)2 = 5 33. (3x - 4)2 = 8 34. (2x + 8)2 = 27

In Exercises 35–46, determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then write and factor the trinomial.

35. x2 + 12x 36. x2 + 16x 37. x2 - 10x 38. x2 - 14x 39. x2 + 3x 40. x2 + 5x 41. x2 - 7x 42. x2 - 9x

43. x2 -23

x 44. x2 +45

x

45. x2 -13

x 46. x2 -14

x

Solve each equation in Exercises 47–64 by completing the square.

47. x2 + 6x = 7 48. x2 + 6x = -8 49. x2 - 2x = 2 50. x2 + 4x = 12 51. x2 - 6x - 11 = 0 52. x2 - 2x - 5 = 0 53. x2 + 4x + 1 = 0 54. x2 + 6x - 5 = 0 55. x2 - 5x + 6 = 0 56. x2 + 7x - 8 = 0 57. x2 + 3x - 1 = 0 58. x2 - 3x - 5 = 0 59. 2x2 - 7x + 3 = 0 60. 2x2 + 5x - 3 = 0 61. 4x2 - 4x - 1 = 0 62. 2x2 - 4x - 1 = 0 63. 3x2 - 2x - 2 = 0 64. 3x2 - 5x - 10 = 0



Solve each equation in Exercises 65–74 using the quadratic formula.

65. x2 + 8x + 15 = 0 66. x2 + 8x + 12 = 0 67. x2 + 5x + 3 = 0 68. x2 + 5x + 2 = 0 69. 3x2 - 3x - 4 = 0 70. 5x2 + x - 2 = 0 71. 4x2 = 2x + 7 72. 3x2 = 6x - 1 73. x2 - 6x + 10 = 0 74. x2 - 2x + 17 = 0

In Exercises 75–82, compute the discriminant. Then determine the number and type of solutions for the given equation.

75. x2 - 4x - 5 = 0 76. 4x2 - 2x + 3 = 0 77. 2x2 - 11x + 3 = 0 78. 2x2 + 11x - 6 = 0 79. x2 - 2x + 1 = 0 80. 3x2 = 2x - 1 81. x2 - 3x - 7 = 0 82. 3x2 + 4x - 2 = 0

Solve each equation in Exercises 83–108 by the method of your choice.

83. 2x2 - x = 1 84. 3x2 - 4x = 4 85. 5x2 + 2 = 11x 86. 5x2 = 6 - 13x 87. 3x2 = 60 88. 2x2 = 250 89. x2 - 2x = 1 90. 2x2 + 3x = 1 91. (2x + 3)(x + 4) = 1 92. (2x - 5)(x + 1) = 2 93. (3x - 4)2 = 16 94. (2x + 7)2 = 25 95. 3x2 - 12x + 12 = 0 96. 9 - 6x + x2 = 0 97. 4x2 - 16 = 0 98. 3x2 - 27 = 0 99. x2 - 6x + 13 = 0 100. x2 - 4x + 29 = 0

101. x2 = 4x - 7 102. 5x2 = 2x - 3 103. 2x2 - 7x = 0 104. 2x2 + 5x = 3

105. 1x

+1

x + 2=

13

106. 1x

+1

x + 3=

14

107. 2x

x - 3+

6x + 3

= - 28

x2 - 9

108. 3

x - 3+

5x - 4

=x2 - 20

x2 - 7x + 12

In Exercises 109–114, fi nd the x@intercept(s) of the graph of each equation. Use the x@intercepts to match the equation with its graph. The graphs are shown in [-10, 10, 1] by [-10, 10, 1] viewing rectangles and labeled (a) through (f).

109. y = x2 - 4x - 5 110. y = x2 - 6x + 7 111. y = -(x + 1)2 + 4 112. y = -(x + 3)2 + 1 113. y = x2 - 2x + 2 114. y = x 2 + 6x + 9

a. b.

c. d.

e. f.

In Exercises 115–122, fi nd all values of x satisfying the given conditions.

115. y = 2x2 - 3x and y = 2. 116. y = 5x2 + 3x and y = 2. 117. y1 = x - 1, y2 = x + 4, and y1y2 = 14. 118. y1 = x - 3, y2 = x + 8, and y1y2 = -30.

119. y1 =2x

x + 2, y2 =

3x + 4

, and y1 + y2 = 1.

120. y1 =3

x - 1, y2 =

8x

, and y1 + y2 = 3.

121. y1 = 2x2 + 5x - 4, y2 = -x2 + 15x - 10, and y1 - y2 = 0.

122. y1 = -x2 + 4x - 2, y2 = -3x2 + x - 1, and y1 - y2 = 0.

Practice Plus In Exercises 123–124, list all numbers that must be excluded from the domain of each rational expression.

123. 3

2x2 + 4x - 9 124.

7

2x2 - 8x + 5

125. When the sum of 6 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

126. When the sum of 1 and twice a negative number is subtracted from twice the square of the number, 0 results. Find the number.

In Exercises 127–130, solve each equation by the method of your choice.

127. 1

x2 - 3x + 2=

1x + 2

+5

x2 - 4

128. x - 1x - 2

+x

x - 3=

1

x2 - 5x + 6

129. 22x2 + 3x - 222 = 0

130. 23x2 + 6x + 723 = 0

Application Exercises In a round-robin chess tournament, each player is paired with every other player once. The formula

N =x2 - x

2

models the number of chess games, N, that must be played in a round-robin tournament with x chess players. Use this formula to solve Exercises 131–132.

131. In a round-robin chess tournament, 21 games were played. How many players were entered in the tournament?

132. In a round-robin chess tournament, 36 games were played. How many players were entered in the tournament?



The graph of the formula in Exercises 131–132 is shown. Use the graph to solve Exercises 133–134.

12

Number of Players

x1 2 3 4 5 6 7 8 9 10 11

Num

ber

of G

ames

Pla

yed

10

20

30

40

50

60

70

N = x2 − x2

N

133. Identify your solution to Exercise 131 as a point on the graph.

134. Identify your solution to Exercise 132 as a point on the graph.

A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died.

Throwing events in track and fi eld include the shot put, the discus throw, the hammer throw, and the javelin throw. The distance that an athlete can achieve depends on the initial velocity of the object thrown and the angle above the horizontal at which the object leaves the hand.

Distance Achieved

Angle at which the shotis released Path of shot

Path's maximum horizontaldistance

In Exercises 137–138, an athlete whose event is the shot put releases the shot with the same initial velocity, but at different angles.

137. When the shot is released at an angle of 35�, its path can be modeled by the formula

y = -0.01x2 + 0.7x + 6.1, in which x is the shot’s horizontal distance, in feet, and y is

its height, in feet. This formula is shown by one of the graphs, (a) or (b), in the fi gure. Use the formula to determine the shot’s maximum distance. Use a calculator and round to the nearest tenth of a foot. Which graph, (a) or (b), shows the shot’s path?

Hei

ght

Horizontal Distance[0, 80, 10] by [0, 40, 10]

(a)

(b)

138. (Refer to the preceding information and the graphs shown in Exercise 137.) When the shot is released at an angle of 65�, its path can be modeled by the formula

y = -0.04x2 + 2.1x + 6.1,

in which x is the shot’s horizontal distance, in feet, and y is its height, in feet. This formula is shown by one of the graphs, (a) or (b), in the fi gure in Exercise 137. Use the formula to determine the shot’s maximum distance. Use a calculator and round to the nearest tenth of a foot. Which graph, (a) or (b), shows the shot’s path?

139. If you have not yet done so, read the Blitzer Bonus on page 159. In this exercise, you will use the golden rectangles shown to obtain an exact value for �, the ratio of the long side to the short side in a golden rectangle of any size.

Golden Rectangle A

Square1 1Golden

RectangleB

�

1 �-1

7975

Fata

l Cra

shes

per

100

Mill

ion

Mile

s D

rive

n

Age of United States Drivers and Fatal Crashes

Age of Drivers6555453525201816

16.3

8.0

3.83.02.42.8

4.1

6.2

9.5

17.718

15

12

9

6

3

0

Source : Insurance Institute for Highway Safety

The number of fatal vehicle crashes per 100 million miles, N, for drivers of age x can be modeled by the formula

N = 0.013x2 - 1.19x + 28.24.

Use the formula to solve Exercises 135–136.

135. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

136. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?



a. The golden ratio in rectangle A, or the ratio of the long

side to the short side, can be modeled by �

1. Write a

fractional expression that models the golden ratio in rectangle B.

b. Set the expression for the golden ratio in rectangle A equal to the expression for the golden ratio in rectangle B. Multiply both sides of the equation by the least common denominator and solve the resulting quadratic equation using the quadratic formula. Express � as an exact value in simplifi ed radical form.

c. Use your solution from part (b) to complete this statement: The ratio of the long side to the short side in a golden rectangle of any size is ____________ to 1.

Use the Pythagorean Theorem and the square root property to solve Exercises 140–143. Express answers in simplifi ed radical form. Then fi nd a decimal approximation to the nearest tenth.

140. A rectangular park is 6 miles long and 3 miles wide. How long is a pedestrian route that runs diagonally across the park?

141. A rectangular park is 4 miles long and 2 miles wide. How long is a pedestrian route that runs diagonally across the park?

142. The base of a 30-foot ladder is 10 feet from a building. If the ladder reaches the fl at roof, how tall is the building?

143. A baseball diamond is actually a square with 90-foot sides. What is the distance from home plate to second base?

144. An isosceles right triangle has legs that are the same length and acute angles each measuring 45�.

a

a

45°

45°

a. Write an expression in terms of a that represents the

length of the hypotenuse. b. Use your result from part (a) to write a sentence that

describes the length of the hypotenuse of an isosceles right triangle in terms of the length of a leg.

145. The length of a rectangular sign is 3 feet longer than the width. If the sign’s area is 54 square feet, fi nd its length and width.

146. A rectangular parking lot has a length that is 3 yards greater than the width. The area of the parking lot is 180 square yards. Find the length and the width.

147. Each side of a square is lengthened by 3 inches. The area of this new, larger square is 64 square inches. Find the length of a side of the original square.

148. Each side of a square is lengthened by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

149. A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width, as shown in the fi gure at the top of the next column. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

10 + 2x

20 + 2xxx

xx

2010

150. A vacant rectangular lot is being turned into a community vegetable garden measuring 15 meters by 12 meters. A path of uniform width is to surround the garden, as shown in the fi gure. If the area of the garden and path combined is 378 square meters, fi nd the width of the path.

xx

15 12

151. A machine produces open boxes using square sheets of metal. The fi gure illustrates that the machine cuts equal-sized squares measuring 2 inches on a side from the corners and then shapes the metal into an open box by turning up the sides. If each box must have a volume of 200 cubic inches, fi nd the length and width of the open box.

x

x

2 2

2

x

2

2

x

152. A machine produces open boxes using square sheets of metal. The machine cuts equal-sized squares measuring 3 inches on a side from the corners and then shapes the metal into an open box by turning up the sides. If each box must have a volume of 75 cubic inches, fi nd the length and width of the open box.

153. A rain gutter is made from sheets of aluminum that are 20 inches wide. As shown in the fi gure, the edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 13 square inches. Show that there are two different solutions to the problem. Round to the nearest tenth of an inch.

20 − 2xx

x

Flat sheet20 inches

wide



154. A piece of wire is 8 inches long. The wire is cut into two pieces and then each piece is bent into a square. Find the length of each piece if the sum of the areas of these squares is to be 2 square inches.

x

cut

8 inches

8 − x

8 − x4

8 − x4

x4

x4

Writing in Mathematics 155. What is a quadratic equation? 156. Explain how to solve x2 + 6x + 8 = 0 using factoring and

the zero-product principle. 157. Explain how to solve x2 + 6x + 8 = 0 by completing the

square. 158. Explain how to solve x2 + 6x + 8 = 0 using the quadratic

formula. 159. How is the quadratic formula derived? 160. What is the discriminant and what information does it

provide about a quadratic equation? 161. If you are given a quadratic equation, how do you determine

which method to use to solve it? 162. Describe the relationship between the real solutions of

ax2 + bx + c = 0 and the graph of y = ax2 + bx + c. 163. If a quadratic equation has imaginary solutions, how is this

shown on the graph of y = ax2 + bx + c?

Technology Exercises 164. Use a graphing utility and x@intercepts to verify any of the

real solutions that you obtained for three of the quadratic equations in Exercises 65–74.

165. Use a graphing utility to graph y = ax2 + bx + c related to any fi ve of the quadratic equations, ax2 + bx + c = 0, in Exercises 75–82. How does each graph illustrate what you determined algebraically using the discriminant?

Critical Thinking Exercises Make Sense? In Exercises 166–169, determine whether each statement makes sense or does not make sense, and explain your reasoning.

166. Because I want to solve 25x2 - 169 = 0 fairly quickly, I’ll use the quadratic formula.

167. I’m looking at a graph with one x@intercept, so it must be the graph of a linear equation.

168. I obtained -17 for the discriminant, so there are two imaginary irrational solutions.

169. When I use the square root property to determine the length of a right triangle’s side, I don’t even bother to list the negative square root.

In Exercises 170–173, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

170. The equation (2x - 3)2 = 25 is equivalent to 2x - 3 = 5. 171. Any quadratic equation that can be solved by completing

the square can be solved by the quadratic formula. 172. The quadratic formula is developed by applying factoring

and the zero-product principle to the quadratic equation ax2 + bx + c = 0.

173. In using the quadratic formula to solve the quadraticequation 5x2 = 2x - 7, we have a = 5, b = 2, and c = -7.

174. Write a quadratic equation in general form whose solution set is {-3, 5}.

175. Solve for t: s = -16t2 + v0t.

176. A rectangular swimming pool is 12 meters long and 8 meters wide. A tile border of uniform width is to be built around the pool using 120 square meters of tile. The tile is from a discontinued stock (so no additional materials are available) and all 120 square meters are to be used. How wide should the border be? Round to the nearest tenth of a meter. If zoning laws require at least a 2-meter-wide border around the pool, can this be done with the available tile?

Preview Exercises Exercises 177–179 will help you prepare for the material covered in the next section.

177. Factor completely: x3 + x2 - 4x - 4. 178. Use the special product (A + B)2 = A2 + 2AB + B2 to

multiply: 11x + 4 + 122.

179. If -8 is substituted for x in the equation 5x

23

+ 11x

13

+ 2 = 0,is the resulting statement true or false?

WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form ax + b = 0, a � 0, and quadratic equations can be written in the general form ax2 + bx + c = 0, a � 0.

We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of ax2 + bx + c = 0, b2 - 4ac, determines the number and

Mid-Chapter Check Point CHAPTER 1


type of solutions. We performed operations with complex numbers and used the imaginary unit i ( i = 2-1, where i2 = -1 ) to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x@intercepts We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions.

In Exercises 1–12, solve each equation.

1. -5 + 3(x + 5) = 2(3x - 4) 2. 5x2 - 2x = 7

3. x - 3

5- 1 =

x - 54

4. 3x2 - 6x - 2 = 0

5. 4x - 2(1 - x) = 3(2x + 1) - 5

6. 5x2 + 1 = 37 7. x(2x - 3) = -4

8. 3x4

-x3

+ 1 =4x5

-3

20 9. (x + 3)2 = 24

10. 1

x2 -4x

+ 1 = 0

11. 3x + 1 - (x - 5) = 2x - 4

12. 2x

x2 + 6x + 8=

xx + 4

-2

x + 2

In Exercises 13–17, fi nd the x-intercepts of the graph of each equation.

13. y = x2 + 6x + 2

14. y = 4(x + 1) - 3x - (6 - x)

15. y = 2x2 + 26

16. y =x2

3+

x2

-23

17. y = x2 - 5x + 8

In Exercises 18–19, fi nd all values of x satisfying the given conditions.

18. y1 = 3(2x - 5) - 2(4x + 1), y2 = -5(x + 3) - 2, and y1 = y2 .

19. y1 = 2x + 3, y2 = x + 2, and y1y2 = 10.

20. Solve by completing the square: x2 + 10x - 3 = 0.

In Exercises 21–22, without solving the equation, determine the number and type of solutions.

21. 2x2 + 5x + 4 = 0 22. 10x(x + 4) = 15x - 15

In Exercises 23–25, graph each equation in a rectangular coordinate system.

23. y = 2x - 1 24. y = 1 - � x �

25. y = x2 + 2

26. Solve for n: L = a + (n - 1)d.

27. Solve for l: A = 2lw + 2lh + 2wh.

28. Solve for f1 : f =f1f2

f1 + f2.

29. What’s the last word in capital punishment? An analysis of the fi nal statements of all men and women Texas has executed since the Supreme Court reinstated the death penalty in 1976 revealed that “love” is by far the most frequently uttered word. The bar graph shows the number of times various words were used in fi nal statements by Texas death-row inmates.

700

500

600

400

300

200

Num

ber

of T

imes

Use

d

Frequently Uttered Words in Final Statements of Death-Row Inmates

Frequently Uttered WordGuiltyPeaceSorryThanksLove

10095

7

Source: Texas Department of Criminal Justice

The number of times “love” was used exceeded the number of times “sorry” was used by 419. The number of utterances of “thanks” exceeded the number of utterances of “sorry” by 32. Combined, these three words were used 1084 times. Determine the number of times each of these words was used in fi nal statements by Texas inmates.

30. The line graph indicates that in 1960, 23% of U.S. taxes came from corporate income tax. For the period from 1960 through 2010, this percentage decreased by approximately 0.28 each year. If this trend continues, by which year will corporations pay zero taxes? Round to the nearest year.

Percentage of U.S. Taxesfrom Corporations

24%

20%

16%

12%

8%

Per

cent

age

of T

axes

from

Cor

pora

tion

s

Year1960 1970 1980 1990 2000 2010

4%

23%

9%

Source: Offi ce of Management and Budget

31. You invested $25,000 in two accounts paying 8% and 9% annual interest. At the end of the year, the total interest from these investments was $2135. How much was invested at each rate?

Mid-Chapter Check Point 165


32. You are choosing between two texting plans. One plan has a monthly fee of $15 with a charge of $0.05 per text. The other plan has a monthly fee of $10 with a charge of $0.075 per text. For how many text messages will the costs for the two plans be the same? What will be the cost for each plan?

33. After a 40% price reduction, you purchase a camcorder for $468. What was the camcorder’s price before the reduction?

34. You invested $4000. On part of this investment, you earned 4% interest. On the remainder of the investment, you lost 3%. Combining earnings and losses, the annual income from these investments was $55. How much was invested at each rate?

In Exercises 35–36, fi nd the dimensions of each rectangle.

35. The rectangle’s length exceeds twice its width by 5 feet. The perimeter is 46 feet.

36. The rectangle’s length is 1 foot shorter than twice its width. The area is 28 square feet.

37. A vertical pole is supported by three wires. Each wire is 13 yards long and is anchored in the ground 5 yards from the base of the pole. How far up the pole will the wires be attached?

38. a. The alligator, at one time an endangered species, was the subject of a protection program. The formula

P = -10x2 + 475x + 3500

models the alligator population, P, after x years of the protection program, where 0 … x … 12. How long did it take the population to reach 5990 alligators?

b. The graph of the formula modeling the alligator population is shown below. Identify your solution from part (a) as a point on the graph.

Alli

gato

r Po

pula

tion

P

x

400045005000550060006500700075008000

3500

Years the Program Is in Effect1 2 3 4 5 6 7 8 9 10 11 12

39. A substantial percentage of the United States population is foreign-born. The bar graph shows the percentage of foreign-born Americans for selected years from 1920 through 2008.

Percentage of the United States PopulationThat Was Foreign-Born, 1920–2008

Year1920

13.2

1930

11.6

1940

8.8

1950

6.9

1960

5.4

1970

4.7

1980

6.2

1990

8.0

2000

10.4

2008

12.5

6%

4%

2%

10%

14%

8%

12%

16%

Per

cent

age

of U

.S. P

opul

atio

n

Source: U.S. Census Bureau

The percentage, p, of the United States population that was foreign-born x years after 1920 can be modeled by the formula

p = 0.004x2 - 0.37x + 14.1. According to this model, in which year will 25% of the United

States population be foreign-born? Round to the nearest year.

In Exercises 40–45, perform the indicated operations and write the result in standard form.

40. (6 - 2i) - (7 - i) 41. 3i(2 + i)

42. (1 + i)(4 - 3i) 43. 1 + i1 - i

44. 2-75 - 2-12 45. 12 - 2-322



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SECTION 1.5 Quadratic Equations - MR. THOMPSON · Section 1.5 Quadratic Equations 147 Solving Quadratic Equations by the Square Root Property Quadratic equations of the form u2 =