section quadratic equations objectives...jan 22, 2016 · quadratic equations of the form where is...

136 Chapter 1 Equations and Inequalities

© 2007 GJ Caulkins

Critical Thinking ExercisesMake Sense? In Exercises 67–70, determine whether eachstatement makes sense or does not make sense, and explain yourreasoning.

67. The humor in this cartoon is based on the fact that “rational”and “real” have different meanings in mathematics and ineveryday speech.

In Exercises 71–74, determine whether each statement is true orfalse. If the statement is false, make the necessary change(s) toproduce a true statement.

71. Some irrational numbers are not complex numbers.

72. is an imaginary number.

73.

74. In the complex number system, (the sum of twosquares) can be factored as

In Exercises 75–77, perform the indicated operations and write theresult in standard form.

75.

76.

77.

Preview ExercisesExercises 78–80 will help you prepare for the material covered inthe next section.

78. Factor:

79. Factor:

80. Evaluate

for and c = -5.a = 2, b = 9,

-b - 2b2 - 4ac2a

x2 - 6x + 9.

2x2 + 7x - 4.

8

1 +2i

1 + i1 + 2i

+1 - i

1 - 2i

412 + i213 - i2

1x + yi21x - yi2.x2 + y2

7 + 3i5 + 3i

=75

13 + 7i213 - 7i2

68. The word imaginary in imaginary numbers tells me that thesenumbers are undefined.

69. By writing the imaginary number I can immediately seethat 5 is the constant and is the variable.

70. When I add or subtract complex numbers, I am basicallycombining like terms.

i5i,

1.5 Quadratic Equations

Do you recognize the significance of this expression? The algebraic expressiongives the solutions of a quadratic equation.In this section, we study a number of methods for solving quadratic equa-

tions. We also look at applications of these equations.

Objectives

1 Solve quadratic equations byfactoring.

2 Solve quadratic equations bythe square root property.

3 Solve quadratic equations bycompleting the square.

4 Solve quadratic equationsusing the quadratic formula.

5 Use the discriminant todetermine the number andtype of solutions.

6 Determine the most efficientmethod to use when solving aquadratic equation.

7 Solve problems modeled byquadratic equations.

S e c t i o n

Definition of a Quadratic EquationA quadratic equation in is an equation that can be written in the general form

where and are real numbers, with A quadratic equation in is alsocalled a second-degree polynomial equation in x.

xa Z 0.ca, b,

ax2 + bx + c = 0,

x

-b 6 2b2 - 4ac2a

M02_BLIT9838_05_SE_01TEST.QXD 7/13/09 9:30 AM Page 136

Check 5: Check 2:

Section 1.5 Quadratic Equations 137

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

We can factor the left side of this equation. We obtain If aquadratic equation has zero on one side and a factored expression on the other side,it can be solved using the zero-product principle.

1x - 521x - 22 = 0.

x2-7x+10=0.

a = 1 b = −7 c = 10

For example, consider the equation According to thezero-product principle, this product can be zero only if at least one of the factors iszero. We set each individual factor equal to zero and solve each resulting equationfor

We can check each of these proposed solutions, 5 and 2, in the original quadraticequation, Substitute each one separately for in the equation.xx2 - 7x + 10 = 0.

x = 5 x = 2 x - 5 = 0 or x - 2 = 01x - 521x - 22 = 0

x.

1x - 521x - 22 = 0.

The resulting true statements indicate that the solutions are 2 and 5.The solution setis Note that with a quadratic equation, we can have two solutions, comparedto the conditional linear equation that had one.52, 56.

� Solve quadratic equations byfactoring.

The Zero-Product PrincipleIf the product of two algebraic expressions is zero, then at least one of the factorsis equal to zero.

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

Solving a Quadratic Equation by Factoring

1. If necessary, rewrite the equation in the general form moving all nonzero terms to one side, thereby obtaining zero on theother side.

2. Factor completely.

3. Apply the zero-product principle, setting each factor containing a variableequal to zero.

4. Solve the equations in step 3.

ax2 + bx + c = 0,

true true 0 = 0,

4 - 14 + 10 � 0

22 - 7 # 2 + 10 � 0 x2 - 7x + 10 = 0

0 = 0,

25 - 35 + 10 � 0

52 - 7 # 5 + 10 � 0 x2 - 7x + 10 = 0

Solving Quadratic Equations by Factoring

Solve by factoring:

a. b. 2x2 + 7x = 4.4x2 - 2x = 0

EXAMPLE 1

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Check 0: Check 12 :

Solution

a. We begin with

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

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

This is the given equation.Factor.

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

Step 5 Check the solutions in the original equation.

x = 12 x = 0 2x = 1 2x = 0 or 2x - 1 = 0

2x12x - 12 = 0 4x2 - 2x = 0

2x

4x2 - 2x = 0.

Check Check �4:12:

true

true 0 = 0, 1 - 1 � 0

4 A14 B - 2 A12 B � 0

4 A12 B2 - 2 A12 B � 0

4x2 - 2x = 0

0 = 0, 0 - 0 � 0

4 # 02 - 2 # 0 � 0 4x2 - 2x = 0

The solution set is

b. Next, we solve

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.

This is the given equation.

Subtract 4 from both sides.

Simplify.

Step 2 Factor.

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

Step 5 Check the solutions in the original equation.

x = 12

2x = 1 x = -4 2x - 1 = 0 or x + 4 = 0 12x - 121x + 42 = 0

2x2 + 7x - 4 = 0

2x2 + 7x - 4 = 0 2x2 + 7x - 4 = 4 - 4

2x2 + 7x = 4

2x2 + 7x = 4.E0, 12 F .

true true 4 = 4, 32 + 1-282 � 4

21-422 + 71-42 � 4 2x2 + 7x = 4

4 = 4, 12 +

72

� 4

2 A12 B2 + 7 A12 B � 4

2x2 + 7x = 4

The solution set is

Check Point 1 Solve by factoring:a. b. 2x2 + x = 1.3x2 - 9x = 0

E -4, 12 F .

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 138


Solving Quadratic Equations by the Square Root PropertyQuadratic equations of the form where is an algebraic expression and isa nonzero real number, can be solved by the square root property. First, isolate thesquared expression on one side of the equation and the number on the otherside. Then take the square root of both sides. Remember, there are two numberswhose square is One number is and one is

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

This is the given equation.Move all terms to one side and obtain zero onthe other side.Factor.

Set each factor equal to zero.

Solve the resulting equations.

Because the solutions differ only in sign, we can write them in abbreviated notationas We read this as “ equals positive or negative the square root of ” or“ equals plus or minus the square root of ”

Now that we have verified these solutions, we can solve directly bytaking square roots. This process is called the square root property.

u2 = dd.u

duu = ; 2d.

u = - 2d u = 2d u + 2d = 0 or u - 2d = 0 Au + 2d B Au - 2d B = 0

u2 - d = 0 u2 = d

u2 = d- 2d .2dd.

du2

duu2 = d,

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

Solving Quadratic Equations by the Square Root Property

Solve by the square root property:

a. b. c.

Solution To apply the square root property, we need a squared expression byitself 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.

1x - 222 = 6.9x2 + 25 = 03x2 - 15 = 0

EXAMPLE 2

� Solve quadratic equations by thesquare root property.

The Square Root PropertyIf is an algebraic expression and is a nonzero real number, then hasexactly two solutions:

Equivalently,

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

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

u2 = ddu

TechnologyGraphic Connections

You can use a graphing utility to check the real solutions of a quadratic equation.The real solutions of correspond to the of the graphof For example, to check the solutions of or

graph The U-shaped, bowl-like, graph isshown on the right. Note that it is important to have all nonzero terms on one sideof the quadratic equation before entering it into the graphing utility. The

are and and the graph of passes throughand This verifies that is the solution set of

or equivalently, 2x2 + 7x = 4.2x2 + 7x - 4 = 0,E -4, 12 FA

12 , 0 B .1-4, 02

y = 2x2 + 7x - 412 ,-4x-intercepts

y = 2x2 + 7x - 4.2x2 + 7x - 4 = 0,2x2 + 7x = 4,y � ax2 � bx � c.

x-interceptsax2 � bx � c � 0

x-intercept is .12x-intercept is −4.

[−5, 2, 1] by [−11, 2, 1]

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The graph of

has no This shows that

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

9x2 + 25 = 0

x-intercepts.

y = 9x2 + 25

[−3, 3, 1] by [−5, 100, 5]


a. This is the original equation.

Add 15 to both sides.

Divide both sides by 3.

Apply the square root property.

Equivalently,

By checking both proposed solutions in the original equation, we can confirmthat the solution set is or

b. This is the original equation.

Subtract 25 from both sides.



Express solutions in terms of

Because the equation has an and no we can check both

proposed solutions, at once.

Check and

true

The solutions are and The solution set is or

c. This is the original equation.


Add 2 to both sides.

By checking both values in the original equation, we can confirm that thesolution set is or

Check Point 2 Solve by the square root property:a. b. c.

Completing the SquareHow do we solve an equation in the form if the trinomial

cannot be factored? We cannot use the zero-product principle in such acase. However, we can convert the equation into an equivalent equation that can besolved using the square root property.This is accomplished by completing the square.

ax2 + bx + cax2 + bx + c = 0

1x + 522 = 11.5x2 + 45 = 03x2 - 21 = 0

E2 ; 26F .E2 + 26, 2 - 26F

x = 2 ; 26

x - 2 = ; 26

1x - 222 = 6

e ; 53

i f .e - 53

i, 53

i f53

i.- 53

i

0 = 0,

25(–1)+25=0?

25i2+25=0?

i2 = −1

9a259

i2b + 25 � 0

9a ; 53

ib2

+ 25 � 0

9x2 + 25 = 0

� 53

i:53

i

; 53

i,

x-term,x2-term

i. x = ; iA

259

= ; 53

i

x = ;A

- 259

x2 = - 259

9x2 = -25 9x2 + 25 = 0

E ; 25F .E - 25, 25F

x = ; 25 . x = 25 or x = - 25

x2 = 5 3x2 = 15

3x2 - 15 = 0

� Solve quadratic equations bycompleting the square.

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 140


xb2

x

b2

Adding this red square

the larger square.

Q Rof area b22

Area of blue region:

Q Rx2 + 2 x = x2 + bxb2

completes

Figure 1.19 Visualizingcompleting the square

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

Completing the Square

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

a. b. c.

Solution To complete the square, we must add a term to each binomial.The termthat should be added is the square of half the coefficient of

a. The coefficient of the in is 8. Half of 8 is 4, and Add16. The result is a perfect square trinomial.

b. The coefficient of the in is Half of is and

Add The result is a perfect square trinomial.

c. The coefficient of the in is Half of is or and

Add The result is a perfect square trinomial.

x2 +35

x +9

100= ax +

310b

2

9100

.a310b

2

=9

100.

310

,12

# 35

,35

35

.x2 +35

xx-term

x2 - 7x +494

= ax -72b

2

494

.a - 72b

2

=494

.

- 72

,-7-7.x2 - 7xx-term

x2 + 8x + 16 = 1x + 422

42 = 16.x2 + 8xx-term

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

35

x.

x2 +35

xx2 - 7xx2 + 8x

EXAMPLE 3

Completing the Square

If is a binomial, then by adding which is the square of half the

coefficient of a perfect square trinomial will result. That is,

x2 + bx + ab

2b

2

= ax +b

2b

2

.

x,

ab

2b

2

,x2 + bx

Study TipYou may not be accustomed to factoring perfect square trinomials in which fractions areinvolved. The constant in the factorization is always half the coefficient of

72

x2-7x+ =ax- b249

472

35

310

Half the coefficient of x, −7, is − . 35310Half the coefficient of x, , is .

x2+ =ax+ b2

x+9

100

x.

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

a. b. c. x2 +23

xx2 - 5xx2 + 6x

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 141

Study TipWhen you complete the square forthe binomial expression you obtain a different polynomial.When you solve a quadratic equationby completing the square, you obtainan equation with the same solutionset because the constant needed tocomplete the square is added to bothsides.

x2 + bx,


We can solve any quadratic equation by completing the square. If thecoefficient of the is one, we add the square of half the coefficient of toboth sides of the equation. When you add a constant term to one side of theequation to complete the square, be certain to add the same constant to the otherside of the equation. These ideas are illustrated in Example 4.

Solving a Quadratic Equation by Completing the Square

Solve by completing the square:

Solution We begin by subtracting 4 from both sides. This is done to isolate thebinomial so that we can complete the square.

This is the original equation.Subtract 4 from both sides.

Next, we work with and complete the square. Find half thecoefficient of the and square it. The coefficient of the is Half of

is and Thus, we add 9 to both sides of the equation.

Add 9 to both sides to complete thesquare.Factor and simplify.


Add 3 to both sides in each equation.

The solutions are and the solution set is or

Check Point 4 Solve by completing the square:

If the coefficient of the in a quadratic equation is not 1, you mustdivide each side of the equation by this coefficient before completing the square.For example, to solve by completing the square, first divide everyterm by 9:

Now that the coefficient of the is 1, we can solve by completing the square.

Solving a Quadratic Equation by Completing the Square

Solve by completing the square:

Solution

This is the original equation.


Add to both sides to isolate the binomial.

Complete the square: Half of is orand

Factor and simplify. ax -13b

2

=59

A� 13 B2 = 19 .�

13 ,

� 26 ,� 23

x2 -23

x +19

=49

+19

49 x

2 -23

x =49

x2 -23

x -49

= 0

9x2 - 6x - 4 = 0

9x2 - 6x - 4 = 0.

EXAMPLE 5

x2-term

x2 -23

x -49

= 0.

x2 -69

x -49

= 0

9x2

9-

6x9

-49

=09

9x2 - 6x - 4 = 0

x2-term

x2 + 4x - 1 = 0.

E3 ; 25F .E3 + 25, 3 - 25F ,3 ; 25

x = 3 + 25 x = 3 - 25 x - 3 = 25 or x - 3 = - 25 1x - 322 = 5

x2 - 6x + 9 = -4 + 9 1-322 = 9.-3-6

-6.x-termx-termx2 - 6x = -4

x2 - 6x = -4 x2 - 6x + 4 = 0

x2 - 6x,

x2 - 6x + 4 = 0.

EXAMPLE 4

xx2-term

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 142



Add to both sides and solve for

Express solutions with a common denominator.

The solutions are and the solution set is

Check Point 5 Solve by completing the square:

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

2x2 + 3x - 4 = 0.

b 1 ; 253

r .1 ; 253

x =1 + 25

3 x = 1 - 25

3

x.13 x =13

+253 x = 1

3-

253

A59

=25

29=

253

x -13

=253 x - 1

3= -

253

x -13

=A

59 or x -

13

= -A

59


Obtain a decimal approximation foreach solution of

the equation in Example 5.

The of verify the solutions.

y = 9x2 - 6x - 4x-intercepts

y = 9x2 − 6x − 4

x-intercept≈ − 0.4

x-intercept≈ 1.1

[−2, 2, 1] by [−10, 10, 1]

1 - 25

3L -0.4

1 + 253

L 1.1;

9x2 - 6x - 4 = 0,

Deriving the Quadratic Formula

General Form of a Quadratic Equation Comment A Specific Example

This is the given equation.

Divide both sides by the coefficient of

x2 +ba

x = - ca

x2 -23

x -43

= 0x2.x2 +ba

x +ca

= 0

3x2 - 2x - 4 = 0ax2 + bx + c = 0, a 7 0

Isolate the binomial by adding on both sidesof the equation.

- ca x

2 -23

x =43

x2 +ba

x +b2

4a2= -

ca

+b2

4a2

x2+b

ax + a

b

2a 2ab

2

= -c

a+ a

bb

2

(half)2

Complete the square. Add the square of half thecoefficient of to both sides.x

x2 -23

x +19

=43

+19

x2-

(half)2

23

x + a-13b

2

=43

+ a-13b

2

ax +b

2ab

2

= - ca

# 4a4a

+b2

4a2Factor on the left side and obtain a commondenominator on the right side. ax -

13b

2

=43

# 33

+19

ax +b

2ab

2

=b2 - 4ac

4a2

ax +b

2ab

2

=-4ac + b2

4a2Add fractions on the right side.

ax -13b

2

=139

ax -13b

2

=12 + 1

9


x +b

2a= ;

3b2 - 4ac2a

x -13

= ;A

139

x +b

2a= ;

B

b2 - 4ac4a2

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

13

= ; 213

3

Solve for by subtracting from both sides.

Combine fractions on the right side. x =1 ; 213

3x =

-b ; 3b2 - 4ac2a

x =13

;213

3b

2axx =

-b2a

;3b2 - 4ac

2a

The formula shown at the bottom of the left column is called the quadraticformula. A similar proof shows that the same formula can be used to solve quadraticequations if the coefficient of the is negative.x2-term,a,

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 143


To use the quadratic formula, rewrite the quadratic equation in general form ifnecessary.Then determine the numerical values for (the coefficient of the ),

(the coefficient of the ), and (the constant term). Substitute the values ofand into the quadratic formula and evaluate the expression. The sign

indicates that there are two (not necessarily distinct) solutions of the equation.

Solving a Quadratic Equation Using the Quadratic Formula

Solve using the quadratic formula:

Solution The given equation is in general form. Begin by identifying the valuesfor and

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

Use the quadratic formula.

Substitute the values for and c:and

and

Complete the subtraction under the radical.

Factor out 2 from the numerator.

Divide the numerator and denominator by 2.

The solution set is or b 3 ; 272

r .b 3 + 272

, 3 - 27

2r

=3 ; 27

2

=2 A3 ; 27 B

4

228 = 24 # 7 = 24 27 = 227 = 6 ; 2274

=6 ; 228

4

4122112 = 8.�1�62 = 6, 1�622 = 1�621�62 = 36,

=6 ; 236 - 8

4

c = 1.a = 2, b = �6,a, b,

=-1-62 ; 41-622 - 4122112

2 # 2

x =-b ; 3b2 - 4ac

2a

2x2-6x+1=0

a = 2 b = −6 c = 1

c.a, b,

2x2 - 6x + 1 = 0.

EXAMPLE 6

;ca, b,cx-termb

x2-terma

The Quadratic FormulaThe solutions of a quadratic equation in general form with

are given by the quadratic formula:

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

divided by 2a.

–b_�b2-4ac2a

x= .

a Z 0,ax2 + bx + c = 0,

TechnologyYou can use a graphing utility to verifythat the solutions of

are Begin by entering

in the screen. Then evaluate this equationat each of the proposed solutions.

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

� Y= �y1 = 2x2 - 6x + 1

3 ; 272

.

2x2 - 6x + 1 = 0

� Solve quadratic equations usingthe quadratic formula.

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 144


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

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

6 ; 227

4=

6 ; 2127

42

=6 ; 27

2 6 ; 227

4=

63

; 22742

=3 ; 227

2

=

1

2

6_2�74

3_�72

2A3_�7B4

= =2A3_�7B

4Factorfirst Then divide

by the GCF.

.

Can all irrational solutions of quadratic equations be simplified? No. Thefollowing solutions cannot be simplified.

Check Point 6 Solve using the quadratic formula:

Solving a Quadratic Equation Using the Quadratic Formula

Solve using the quadratic formula:Solution The given equation is in general form. Begin by identifying the valuesfor and


Substitute the values for and c:and

and

Subtract under the radical. Because the numberunder the radical sign is negative, the solutions willnot be real numbers.

Factor 2 from the numerator.

Divide the numerator and denominator by 2.

Write the complex numbers in standard form.

The solutions are complex conjugates, and the solution set is

or b 13

; i 211

3r .1

3- i

2113rb 1

3+ i

2113

,

=13

; i 211

3

=1 ; i211

3

=2 A1 ; i211 B

6

2�44 = 4411121� 12 = 2i211 =2 ; 2i211

6

=2 ; 2-44

6

1�222 = 1�221�22 = 4.�1�22 = 2 =2 ; 24 - 48

6

c = 4.a = 3, b = �2,a, b,

=-1-22 ; 41-222 - 4132142

2132

x =-b ; 3b2 - 4ac

2a

3x2-2x+4=0

a = 3 b = −2 c = 4

c.a, b,

3x2 - 2x + 4 = 0.

EXAMPLE 7

2x2 + 2x - 1 = 0.

5_2�72

–4_3�72

Other than 1, terms ineach numerator have no

common factor.

Study TipChecking irrational and imaginarysolutions can be time-consuming.The solutions given by the quadraticformula are always correct, unlessyou have made a careless error.Checking for computational errorsor errors in simplification issufficient.

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If has imaginary solutions,the graph of will not have

This is illustrated by the imaginarysolutions of in Example 7 andthe graph of in Figure 1.20.

Check Point 7 Solve using the quadratic formula:

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

b2 - 4ac,

x2 - 2x + 2 = 0.

y = 3x2 - 2x + 43x2 - 2x + 4 = 0

x-intercepts.y = ax2 + bx + c

ax2 + bx + c = 0

� Use the discriminant todetermine the number and typeof solutions.

y = 3x2 − 2x + 4

[−2, 2, 1] by [−1, 10, 1]

Figure 1.20 This graph has no.x-intercepts

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

Discriminantb2 � 4ac

Kinds of Solutions to ax2 � bx � c � 0

Graph ofy � ax2 � bx � c

b2 - 4ac 7 0 Two unequal real solutions:If and are rational numbers and thediscriminant is a perfect square, thesolutions are rational. If the discriminant isnot a perfect square, the solutions areirrational conjugates.

ca, b,

Two -interceptsx

x

y

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

ca, b,

One -interceptx

x

y

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

No -interceptsx

x

y

Using the Discriminant

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

a. b. c.

Solution Begin by identifying the values for and in each equation. Thencompute the discriminant.

a. 3x2+4x-5=0

a = 3 b = 4 c = −5

b2 - 4ac,ca, b,

3x2 - 8x + 7 = 0.9x2 - 6x + 1 = 03x2 + 4x - 5 = 0

EXAMPLE 8

Study Tip

The discriminant is It is not so do not give the

discriminant as a radical.3b2 - 4ac ,

b2 - 4ac.

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 146


Substitute and compute the discriminant:

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

Substitute and compute the discriminant:

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

The negative discriminant, 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 determinethe number and type of solutions:

a. b. c.

Determining Which Method to UseAll quadratic equations can be solved by the quadratic formula. However, if anequation is in the form such as or it is fasterto use the square root property, taking the square root of both sides. If the equationis not in the form write the quadratic equation ingeneral form Try to solve the equation by factoring. If

cannot be factored, then solve the quadratic equation by thequadratic formula.

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

ax2 + bx + c1ax2 + bx + c = 02.

u2 = d,

12x + 322 = 8,x2 = 5u2 = d,

3x2 - 2x + 4 = 0.2x2 - 7x - 4 = 0x2 + 6x + 9 = 0

-20,

b2 - 4ac = 1-822 - 4 # 3 # 7 = 64 - 84 = -20

c. 3x2-8x+7=0

a = 3 c = 7b = −8

b2 - 4ac = 1-622 - 4 # 9 # 1 = 36 - 36 = 0.

b. 9x2-6x+1=0

a = 9 b = −6 c = 1

b2 - 4ac = 42 - 4 # 31-52 = 16 - 1-602 = 16 + 60 = 76.

� Determine the most efficientmethod to use when solving aquadratic equation.

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 147


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

Description and Form of the Quadratic Equation

Most Efficient Solution Method Example

and can be factored easily.

ax2 + bx + cax2 + bx + c = 0 Factor and use the zero-product principle.

x =13 x = -2

3x - 1 = 0 or x + 2 = 0 13x - 121x + 22 = 0 3x2 + 5x - 2 = 0

The quadratic equation has no constantterm. 1c = 02

ax2 + bx = 0 Factor and use the zero-product principle.

x = - 32

x = 0 2x = -3 3x = 0 or 2x + 3 = 0 3x12x + 32 = 0

6x2 + 9x = 0

The quadratic equation has no 1b = 02

x-term.ax2 + c = 0 Solve for and apply

the square root property.x2

= ; 2

27= ;

2

27# 2727

= ; 227

7

x = ;A

47

x2 =47

7x2 = 4 7x2 - 4 = 0

is a first-degree polynomial.u2 = d; u Use the square root property.

x = -4 ; 25

x + 4 = ; 25

1x + 422 = 5

and cannot be factored or the factoring is too difficult.

ax2 + bx + cax2 + bx + c = 0 Use the quadratic formula:

x =-b ; 3b2 - 4ac

2a.

= 1 ; 27

=2 ; 227

2=

2 A1 ; 27 B2

=2 ; 228

2=

2 ; 24272

=2 ; 44 + 24

2112

x =-1-22 ; 41-222 - 41121-62

2112

x¤-2x-6=0

a = 1 b = −2 c = −6

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

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 148


Applications

Blood Pressure and Age

The graphs in Figure 1.21 illustrate that aperson’s normal systolic blood pressure,measured in millimeters of mercury (mmHg), depends on his or her age. (Thesymbol on the vertical axis showsthere is a break in values between 0 and100. Thus, the first tick mark on thevertical axis represents a blood pressureof 100 mm Hg.) The formula

models a man’s normal systolic pressure,at age

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

b. Use the graphs in Figure 1.21 to describe the differences between the normalsystolic 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 of125 millimeters of mercury. Thus, we substitute 125 for in the given formulafor men. Then we solve for 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

A,P

A.P,

P = 0.006A2 - 0.02A + 120

EXAMPLE 9

� Solve problems modeled byquadratic 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

Subtract 125 from both sides andwrite the quadratic equation ingeneral form.

P.

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

Notice that thevariable is A,rather than the

usual x.A=

–b_�b2-4ac2a

=–(–0.02)_�(–0.02)2-4(0.006)(–5)

2(0.006)


Substitute the valuesfor and

andc = �5.b = �0.02,a = 0.006,

c:a, b,

Use a calculator tosimplify the radicand.

Use a calculator:

Use a calculator andround to the nearestinteger.Reject this solution.

Age cannot be negative.

A≠ –27 A L 31

A L0.02 + 0.347

0.012 or A L 0.02 - 0.347

0.012

20.1204 « 0.347. L0.02 + 0.347

0.012

=0.02 ; 20.1204

0.012

TechnologyOn most calculators, here is how toapproximate

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 )

0.02 + 20.12040.012

.

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 149


The positive solution,indicates that 31 is the approximateage of a man whose normal systolicblood pressure is 125 mm Hg. Thisis illustrated by the black lines withthe arrows on the red graphrepresenting men in Figure 1.22.

b. Take a second look at the graphs inFigure 1.22. Before approximatelyage 50, the blue graph representingwomen’s normal systolic bloodpressure lies below the red graphrepresenting men’s normal systolicblood pressure. Thus, up to age 50,women’s normal systolic bloodpressure is lower than men’s, although it is increasing at a faster rate. After age50, women’s normal systolic blood pressure is higher than men’s.

Check Point 9 The formula models a woman’snormal systolic blood pressure, at age Use this formula to find the age, tothe 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 amathematical 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 oneangle measuring 90°.The side opposite the 90° angle is called the hypotenuse.The othersides 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 squaresis 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 aright triangle and is called the Pythagorean Theorem.

A.P,P = 0.01A2 + 0.05A + 107

A L 31,

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 thelarge square equals the sum of theareas of the smaller squares.

The Pythagorean TheoremThe sum of the squares of the lengths of the legs of aright triangle equals the square of the length of thehypotenuse.

If the legs have lengths and and thehypotenuse has length then

a2 + b2 = c2.

c,b,a

Using the Pythagorean Theorem

a. A wheelchair ramp with a length of 122 inches has a horizontal distance of120 inches. What is the ramp’s vertical distance?

b. Construction laws are very specific when it comes to access ramps for thedisabled. Every vertical rise of 1 inch requires a horizontal run of 12 inches.Does this ramp satisfy the requirement?

Solutiona. Figure 1.24 shows the right triangle that is formed by the ramp, the wall, and the

ground.We can find the ramp’s vertical distance,using the Pythagorean Theorem.

x2 + 1202 = 1222

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

x,

EXAMPLE 10

120 in.

x 122 in.

Figure 1.24

Hypotenusec

B

CA bLeg

aLeg

C-BLTZMC01_089-196-hr2 12-09-2008 10:39 Page 150


We solve this equation using the square root property.

This is the equation resulting from the PythagoreanTheorem.Square 120 and 122.Isolate by subtracting 14,400 from both sides.Apply the square root property.

Because 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 theramp has a vertical distance of 22 inches, it requires a horizontal distance of22(12) inches, or 264 inches. The horizontal distance is only 120 inches, so thisramp does not satisfy construction laws for access ramps for the disabled.

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

x

x = 22 x = -22 x = 2484 or x = - 2484x2 x2 = 484

x2 + 14,400 = 14,884

x2 + 1202 = 1222

Art, Nature, and Quadratic Equations

In The Bathers at Asnieres, by the French impressionistGeorges Seurat (1859–1891), the artist positions parts of the painting as though they were inside golden rectangles.

Georges Seurat, “The Bathers at Asnieres,” 1883–1884. Oil oncanvas, Erich Lessing/Art Resource, NY.79 12 * 118

12 in.

Golden Rectangle A

SquareGolden

RectangleB

Figure 1.25(a)

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

£ L 1.6.£

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. Ifthe smaller golden rectangle is divided again, the same is true of the yet smaller rectangle, and so on. The process of repeatedlydividing each golden rectangle in this manner is illustrated in Figure 1.25(b). We’ve also created a spiral by connecting the oppositecorners of all the squares with a smooth curve. This spiral matches the spiral shape of the chambered nautilus shell shown in Figure1.25(c). The shell spirals out at an ever-increasing rate that is governed by this geometry.

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 ofthe long side to the short side in a golden rectangle of any size. Your model will involve a quadratic equation that can be solved bythe quadratic formula. (See Exercise 137.)

£,

Figure 1.25(b) Figure 1.25(c)

C-BLTZMC01_089-196-hr2 12-09-2008 10:40 Page 151


Exercise Set 1.5

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

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

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

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

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

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

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

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

57. 58.

59. 60.

61. 62.

63. 64.

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

65. 66.

67. 68.

69. 70.

71. 72.

73. 74. x2 - 2x + 17 = 0x2 - 6x + 10 = 03x2 = 6x - 14x2 = 2x + 75x2 + x - 2 = 03x2 - 3x - 4 = 0x2 + 5x + 2 = 0x2 + 5x + 3 = 0x2 + 8x + 12 = 0x2 + 8x + 15 = 0

3x2 - 5x - 10 = 03x2 - 2x - 2 = 02x2 - 4x - 1 = 04x2 - 4x - 1 = 02x2 + 5x - 3 = 02x2 - 7x + 3 = 0x2 - 3x - 5 = 0x2 + 3x - 1 = 0x2 + 7x - 8 = 0x2 - 5x + 6 = 0x2 + 6x - 5 = 0x2 + 4x + 1 = 0x2 - 2x - 5 = 0x2 - 6x - 11 = 0x2 + 4x = 12x2 - 2x = 2x2 + 6x = -8x2 + 6x = 7

x2 -14

xx2 -13

x

x2 +45

xx2 -23

x

x2 - 9xx2 - 7xx2 + 5xx2 + 3xx2 - 14xx2 - 10xx2 + 16xx2 + 12x

12x + 822 = 2713x - 422 = 818x - 322 = 515x - 122 = 714x - 122 = 1613x + 222 = 91x + 222 = -71x - 322 = -51x - 122 = -91x + 322 = -1631x + 422 = 2131x - 422 = 151x - 322 = 361x + 222 = 252x2 - 7 = -152x2 - 5 = -553x2 - 1 = 475x2 + 1 = 515x2 = 453x2 = 27

10x - 1 = 12x + 1227 - 7x = 13x + 221x - 1216x1x - 22 = 8x - 252x1x - 32 = 5x2 - 7x5x2 - 20x = 03x2 + 12x = 04x2 - 13x = -33x2 - 2x = 89x2 + 9x + 2 = 06x2 + 11x - 10 = 0x2 = -11x - 10x2 = 8x - 15x2 - 13x + 36 = 0x2 - 3x - 10 = 0

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

75. 76.

77. 78.

79. 80.

81. 82.

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

83. 84.

85. 86.

87. 88.

89. 90.

91. 92.

93. 94.

95. 96.

97. 98.

99. 100.

101. 102.

103. 104.

105. 106.

107.

108.

In Exercises 109–114, find the of the graph of eachequation. Use the to match the equation with itsgraph. The graphs are shown in by viewing rectangles and labeled through

109. 110.

111. 112.

113. 114.

a. b.

c. d.

e. f.

y = x2 + 6x + 9y = x2 - 2x + 2y = -1x + 322 + 1y = -1x + 122 + 4y = x2 - 6x + 7y = x2 - 4x - 51f2.1a2

3-10, 10, 143-10, 10, 14x-intercepts

x-intercept1s2

3x - 3

+5

x - 4=

x2 - 20x2 - 7x + 12

2xx - 3

+6

x + 3= -

28x2 - 9

1x

+1

x + 3=

14

1x

+1

x + 2=

13

2x2 + 5x = 32x2 - 7x = 05x2 = 2x - 3x2 = 4x - 7x2 - 4x + 29 = 0x2 - 6x + 13 = 03x2 - 27 = 04x2 - 16 = 09 - 6x + x2 = 03x2 - 12x + 12 = 012x + 722 = 2513x - 422 = 1612x - 521x + 12 = 212x + 321x + 42 = 12x2 + 3x = 1x2 - 2x = 12x2 = 2503x2 = 605x2 = 6 - 13x5x2 + 2 = 11x3x2 - 4x = 42x2 - x = 1

3x2 + 4x - 2 = 0x2 - 3x - 7 = 0

3x2 = 2x - 1x2 - 2x + 1 = 0

2x2 + 11x - 6 = 02x2 - 11x + 3 = 0

4x2 - 2x + 3 = 0x2 - 4x - 5 = 0

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In Exercises 115–122, find all values of satisfying the givenconditions.

115. and

116. and

117. and

118. and

119. and

120. and

121. and

122. and

Practice PlusIn Exercises 123–124, list all numbers that must be excluded fromthe domain of each rational expression.

123. 124.

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

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

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

127.

128.

129.

130.

Application ExercisesIn a round-robin chess tournament, each player is paired withevery other player once. The formula

models the number of chess games, that must be played in around-robin tournament with chess players. Use this formula tosolve 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?

xN,

N =x2 - x

2

23x2 + 6x + 723 = 0

22x2 + 3x - 222 = 0

x - 1x - 2

+x

x - 3=

1x2 - 5x + 6

1x2 - 3x + 2

=1

x + 2+

5x2 - 4

72x2 - 8x + 5

32x2 + 4x - 9

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

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

y1 + y2 = 3.y1 =3

x - 1, y2 =

8x

,

y1 + y2 = 1.y1 =2x

x + 2, y2 =

3x + 4

,

y1y2 = -30.y1 = x - 3, y2 = x + 8,

y1y2 = 14.y1 = x - 1, y2 = x + 4,

y = 2.y = 5x2 + 3x

y = 2.y = 2x2 - 3x

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

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

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

Throwing events in track and field include the shot put, the discusthrow, the hammer throw, and the javelin throw. The distance thatan athlete can achieve depends on the initial velocity of the objectthrown and the angle above the horizontal at which the objectleaves the hand.

In Exercises 135–136, an athlete whose event is the shot put releasesthe shot with the same initial velocity, but at different angles.

135. When the shot is released at an angle of 35°, its path can bemodeled by the formula

in which is the shot’s horizontal distance, in feet, and isits height, in feet. This formula is shown by one of thegraphs, (a) or (b), in the figure. Use the formula to deter-mine the shot’s maximum distance. Use a calculator andround 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)

yx

y = -0.01x2 + 0.7x + 6.1,

Distance Achieved

Angle at which the shotis released Path of shot

Path's maximum horizontaldistance

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

C-BLTZMC01_089-196-hr2 12-09-2008 10:40 Page 153


136. (Refer to the information on the previous page and thegraphs shown in Exercise 135.) When the shot is released atan angle of 65°, its path can be modeled by the formula

in which is the shot’s horizontal distance, in feet, and isits height, in feet. This formula is shown by one of thegraphs, (a) or (b), in the figure in Exercise 135. Use theformula to determine the shot’s maximum distance. Use acalculator and round to the nearest tenth of a foot. Whichgraph, (a) or (b), shows the shot’s path?

137. If you have not yet done so, read the essay on page 151. Inthis exercise, you will use the golden rectangles shown toobtain an exact value for the ratio of the long side to theshort side in a golden rectangle of any size.

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

side to the short side, can be modeled by Write a

fractional expression that models the golden ratio inrectangle

b. Set the expression for the golden ratio in rectangle equal to the expression for the golden ratio in rectangle

Multiply both sides of the equation by the least com-mon denominator and solve the resulting quadraticequation using the quadratic formula. Express as anexact value in simplified radical form.

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

Use the Pythagorean Theorem and the square root property tosolve Exercises 138–141. Express answers in simplified radicalform. Then find a decimal approximation to the nearest tenth.

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

139. A rectangular park is 4 miles long and 2 miles wide. Howlong is a pedestrian route that runs diagonally across thepark?

140. The base of a 30-foot ladder is 10 feet from a building. If theladder reaches the flat roof, how tall is the building?

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

142. An isosceles right triangle has legs that are the same lengthand acute angles each measuring 45°.

a

a

45°

45°

£

B.

A

B.

£1

.

A,

Golden Rectangle A

Square1 1Golden

RectangleB

�

1 �-1

£,

yx

y = -0.04x2 + 2.1x + 6.1,

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

b. Use your result from part (a) to write a sentence thatdescribes the length of the hypotenuse of an isoscelesright triangle in terms of the length of a leg.

143. The length of a rectangular sign is 3 feet longer than thewidth. If the sign’s area is 54 square feet, find its length andwidth.

144. A rectangular parking lot has a length that is 3 yards greaterthan the width. The area of the parking lot is 180 squareyards. Find the length and the width.

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

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

147. A pool measuring 10 meters by 20 meters is surrounded bya path of uniform width, as shown in the figure. If the areaof the pool and the path combined is 600 square meters,what is the width of the path?

148. A vacant rectangular lot is being turned into a communityvegetable garden measuring 15 meters by 12 meters. A pathof uniform width is to surround the garden, as shown in thefigure. If the area of the garden and path combined is378 square meters, find the width of the path.

149. A machine produces open boxes using square sheets ofmetal. The figure illustrates that the machine cuts equal-sized squares measuring 2 inches on a side from the cornersand then shapes the metal into an open box by turning upthe sides. If each box must have a volume of 200 cubicinches, find the length and width of the open box.

x

x

2 2

2

x

2

2

x

xx

15 12

10 + 2x

20 + 2xxx

xx

2010

a

C-BLTZMC01_089-196-hr2 12-09-2008 10:40 Page 154


150. A machine produces open boxes using square sheets ofmetal. The machine cuts equal-sized squares measuring3 inches on a side from the corners and then shapes themetal into an open box by turning up the sides. If each boxmust have a volume of 75 cubic inches, find the length andwidth of the open box.

151. A rain gutter is made from sheets of aluminum that are20 inches wide. As shown in the figure, the edges are turnedup to form right angles. Determine the depth of the gutterthat 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.

152. A piece of wire is 8 inches long. The wire is cut into twopieces and then each piece is bent into a square. Find thelength of each piece if the sum of the areas of these squaresis to be 2 square inches.

Writing in Mathematics153. What is a quadratic equation?

154. Explain how to solve using factoring andthe zero-product principle.

155. Explain how to solve by completing thesquare.

156. Explain how to solve using the quadraticformula.

157. How is the quadratic formula derived?

158. What is the discriminant and what information does itprovide about a quadratic equation?

159. If you are given a quadratic equation, how do youdetermine which method to use to solve it?

160. Describe the relationship between the real solutions ofand the graph of

161. If a quadratic equation has imaginary solutions, how is thisshown on the graph of y = ax2 + bx + c?

y = ax2 + bx + c.ax2 + bx + c = 0

x2 + 6x + 8 = 0

x2 + 6x + 8 = 0

x2 + 6x + 8 = 0

x

cut

8 inches

8 − x

8 − x4

8 − x4

x4

x4

20 − 2xxx

Flat sheet20 inches

wide

Technology Exercises162. Use a graphing utility and to verify any of the

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

163. Use a graphing utility to graph related toany five of the quadratic equations, inExercises 75–82. How does each graph illustrate what youdetermined algebraically using the discriminant?

Critical Thinking Exercises

Make Sense? In Exercises 164–167, determine whether eachstatement makes sense or does not make sense, and explain yourreasoning.

164. Because I want to solve fairly quickly, I’lluse the quadratic formula.

165. I’m looking at a graph with one so it must bethe graph of a linear equation.

166. I obtained for the discriminant, so there are twoimaginary irrational solutions.

167. When I use the square root property to determine thelength of a right triangle’s side, I don’t even bother to listthe negative square root.

In Exercises 168–171, determine whether each statement is true orfalse. If the statement is false, make the necessary change(s) toproduce a true statement.

168. The equation is equivalent to

169. Any quadratic equation that can be solved by completingthe square can be solved by the quadratic formula.

170. The quadratic formula is developed by applying factoringand the zero-product principle to the quadratic equation

171. In using the quadratic formula to solve the quadraticequation we have and

172. Write a quadratic equation in general form whose solutionset is

173. Solve for

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

Preview ExercisesExercises 175–177 will help you prepare for the material coveredin the next section.

175. Factor completely:

176. Use the special product tomultiply:

177. If is substituted for in the equation is the resulting statement true or false?

5x 23

+ 11x 13

+ 2 = 0,x-8

A2x + 4 + 1 B2.1A + B22 = A2 + 2AB + B2

x3 + x2 - 4x - 4.

t: s = -16t2 + v0t.

5-3, 56.

c = -7.a = 5, b = 2,5x2 = 2x - 7,

ax2 + bx + c = 0.

2x - 3 = 5.12x - 322 = 25

-17

x-intercept,

25x2 - 169 = 0

ax2 + bx + c = 0,y = ax2 + bx + c

x-intercepts

C-BLTZMC01_089-196-hr2 12-09-2008 10:40 Page 155

section quadratic equations objectives...jan 22, 2016 · quadratic equations of the form where is...

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