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D. The Quadratic Formula and the Discriminant

# D. The Quadratic Formula and the Discriminant

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On the left, our final solutions are x ϭ Ϫ1 or x ϭ Ϫ32. The general solution is

called the quadratic formula, which can be used to solve any equation belonging to

If ax2 ϩ bx ϩ c ϭ 0, with a, b, and c ʦ ‫ ޒ‬and a 0, then

Ϫb ؊ 2b2 Ϫ 4ac

Ϫb ؉ 2b2 Ϫ 4ac

or

;

2a

2a

Ϫb ؎ 2b2 Ϫ 4ac

.

also written x ϭ

2a

CAUTION

EXAMPLE 5

It’s very important to note the values of a, b, and c come from an equation written in standard form. For 3x2 Ϫ 5x ϭ Ϫ7, a ϭ 3 and b ϭ Ϫ5, but c Ϫ7! In standard form we have

3x2 Ϫ 5x ϩ 7 ϭ 0, and note the value for use in the formula is actually c ϭ 7.

Solve 4x2 ϩ 1 ϭ 8x using the quadratic formula. State the solution(s) in both exact

and approximate form. Check one of the exact solutions in the original equation.

Solution

Begin by writing the equation in standard form and identifying the values of a, b,

and c.

4x2 ϩ 1 ϭ 8x

4x Ϫ 8x ϩ 1 ϭ 0

a ϭ 4, b ϭ Ϫ8, c ϭ 1

2

Ϫ1Ϫ82 Ϯ 21Ϫ82 2 Ϫ 4142112

2142

8 Ϯ 148

8 Ϯ 164 Ϫ 16

ϭ

8

8

8 Ϯ 4 13

8

4 13

ϭ Ϯ

8

8

8

13

13

xϭ1ϩ

or x ϭ 1 Ϫ

2

2

or x Ϸ 0.13

x Ϸ 1.87

Check

4x2 ϩ 1 ϭ 8x

13 2

13

4 a1 ϩ

b ϩ 1 ϭ 8 a1 ϩ

b

2

2

4 c 1 ϩ 2a

13

3

b ϩ d ϩ 1 ϭ 8 ϩ 4 13

2

4

4 ϩ 4 13 ϩ 3 ϩ 1 ϭ 8 ϩ 4 13

8 ϩ 4 13 ϭ 8 ϩ 4 13 ✓

original equation

standard form

substitute 4 for a, ؊8 for b, and 1 for c

simplify

exact solutions

approximate solutions

original equation

substitute 1 ϩ

13

2

for x

square binomial; distribute

distribute

result checks

A graphing calculator check is also shown.

Now try Exercises 61 through 90

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1

CAUTION

For

8 Ϯ 413

8 Ϯ 413

ϭ 1 Ϯ 413.

, be careful not to incorrectly “cancel the eights” as in

8

8

1

No! Use a calculator to verify that the results are not equivalent. Both terms in the numerator are divided by 8 and we must either rewrite the expression as separate terms (as

above) or factor the numerator to see if the expression simplifies further:

1

8 Ϯ 413

2 Ϯ 13

4 12 Ϯ 132

13

ϭ

ϭ

, which is equivalent to 1 Ϯ

.

8

8

2

2

2

The Discriminant of the Quadratic Formula

The conclusions we reached graphically in Figure 3.14 regarding the nature of number

of quadratic roots can now be seen algebraically through a closer look at the quadratic

formula. For any real-valued expression X, recall that 1X represents a real number

only for X Ն 0. Since the quadratic formula contains the radical 2b2 Ϫ 4ac, the

expression b2 Ϫ 4ac, called the discriminant, will determine the nature (real or complex) and the number of solutions to a given quadratic equation.

The Discriminant of the Quadratic Formula

For f 1x2 ϭ ax2 ϩ bx ϩ c, where a, b, c ʦ ‫ ޒ‬and a

1. If b Ϫ 4ac 7 0,

there are two

real roots

2

1. f 1x2 ϭ x2 Ϫ 4x

a ϭ 1, b ϭ Ϫ4, c ϭ 0

b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112102

ϭ 16 Ϫ 0

ϭ 16

16 7 0, f has two real roots

0,

3. If b2 Ϫ 4ac 6 0,

there are two

nonreal roots

These ideas are further illustrated here, by comparing the value of the discriminant

with the graph of the related function shown below each calculation.

2. g1x2 ϭ x2 Ϫ 4x ϩ 4

a ϭ 1, b ϭ Ϫ4, c ϭ 4

b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112142

ϭ 16 Ϫ 16

ϭ0

0 ϭ 0, g has one real root

10

Ϫ5

2. If b Ϫ 4ac ϭ 0,

there is one real

(repeated) root

2

3. h1x2 ϭ x2 Ϫ 4x ϩ 5

a ϭ 1, b ϭ Ϫ4, c ϭ 5

b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112152

ϭ 16 Ϫ 20

ϭ Ϫ4

Ϫ4 6 0, h has two nonreal roots

10

10

Ϫ5

10

10

Ϫ5

Ϫ5

Ϫ5

10

Ϫ5

Finally, we note from (3) and the structure of the quadratic formula, that the

complex solutions must occur in conjugate pairs.

Complex Solutions

The complex solutions of a quadratic equation with real coefficients

must occur in conjugate pairs.

If a ϩ bi is a solution, then a Ϫ bi is also a solution.

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Further analysis of the discriminant reveals even more concerning the nature of

quadratic solutions. Namely, if a, b, and c are rational and the discriminant is

1. zero, the original equation is a perfect square trinomial.

2. a perfect square, there will be two rational roots which means

the original equation can be solved by factoring.

3. not a perfect square, there will be two irrational roots.

See Exercises 91 through 102.

EXAMPLE 6

Solve: 2x2 Ϫ 6x ϩ 5 ϭ 0.

Solution

With a ϭ 2, b ϭ Ϫ6, and c ϭ 5, the discriminant becomes 1Ϫ62 2 Ϫ 4122152 ϭ Ϫ4,

showing there will be two complex roots. The quadratic formula then yields

Ϫb Ϯ 2b2 Ϫ 4ac

2a

Ϫ1Ϫ62 Ϯ 1Ϫ4

2122

6 Ϯ 2i

4

3

1

xϭ Ϯ i

2

2

b 2 Ϫ 4ac ϭ Ϫ4, substitute 2 for a and ؊6 for b

simplify, write in i form

solutions are complex conjugates

A calculator check is shown in Figures 3.25 and 3.26.

Figure 3.25, 3.26

WORTHY OF NOTE

While it’s possible to solve by

b

completing the square if is a

a

fraction or an odd number (see

Example 4), the process is usually

b

most efficient when is an even

a

number. This is one observation

you could use when selecting a

solution method.

D. You’ve just seen how

formula and the discriminant

Now try Exercises 103 through 108

Summary of Solution Methods for ax2 ؉ bx ؉ c ‫ ؍‬0

1. If b ϭ 0, ax2 ϩ c ϭ 0: isolate x2 and use the square root property of equality.

2. If c ϭ 0, ax2 ϩ bx ϭ 0: factor out the GCF and use the zero product property.

3. If no coefficient is zero, you can attempt to solve by

a. factoring b. completing the square c. using the quadratic formula

d. using the intersection-of-graphs method e. using the zeroes method

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Section 3.2 Solving Quadratic Equations and Inequalities

The study of quadratic inequalities is simply an extension of our earlier work in analyzing functions (Section 2.1). While we’ve developed the ability to graph a variety of

new functions, the solution set for an inequality will still be determined by analyzing

the behavior of the function at its zeroes. The key idea is to recognize the following

statements are synonymous:

1. f 1x2 7 0.

2. Outputs are positive.

3. The graph is above the x-axis.

Similar statements can be made using the other inequality symbols.

Solving a quadratic inequality only requires that we (a) locate any real zeroes of

the function and (b) determine whether the graph opens upward or downward. If there

are no x-intercepts, the graph is entirely above the x-axis (output values are positive),

or entirely below the x-axis (output values are negative), making the solution either all

real numbers or the empty set.

EXAMPLE 7

Analytical Solution

For f 1x2 ϭ x2 ϩ x Ϫ 6, solve f 1x2 7 0.

The graph of f will open upward since a 7 0. Factoring gives f 1x2 ϭ 1x ϩ 321x Ϫ 22 ,

with zeroes at Ϫ3 and 2. Using the x-axis alone (since graphing the function is not

our focus), we plot (Ϫ3, 0) and (2, 0) and visualize a parabola opening upward

through these points (Figure 3.27).

Figure 3.27

When Ϫ3 Ͻ x Ͻ 2,

the graph is below

the x-axis,

f(x) Ͻ 0.

Ϫ5

Ϫ4

Ϫ3

Ϫ2

Ϫ1

0

aϾ0

1

2

x

When x Ͼ 2,

the graph is above

the x-axis,

f(x) Ͼ 0.

When x Ͻ Ϫ3,

the graph is above

the x-axis,

f(x) Ͼ 0.

The diagram clearly shows the graph is above the x-axis (outputs are positive)

when x 6 Ϫ3 or when x 7 2. The solution is x ʦ 1Ϫq, Ϫ32 ´ 12, q 2 .

Graphical Solution

The complete graph of f shown in Figure 3.28

confirms the analytical solution. For the intervals

of the domain shown in bold 1Ϫq, Ϫ32 ´ 12, q 2 ,

the graph is above the x-axis 1 f 1x2 7 02 . For the

portion shown in red, the graph is below the

x-axis 1 f 1x2 6 02 .

Figure 3.28

5

Ϫ5

y f(x) ϭ x2 ϩ x Ϫ 6

5

x

Ϫ5

Now try Exercises 109 through 120

When solving general inequalities, zeroes of multiplicity continue to play a role.

In Example 7, the zeroes of f were both of multiplicity 1, and the graph crossed the

x-axis at these points. In other cases, the zeroes may have even multiplicity.

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EXAMPLE 8

Solve the inequality Ϫx2 ϩ 6x Յ 9.

Analytical Solution

WORTHY OF NOTE

Begin by writing the inequality in standard form: Ϫx2 ϩ 6x Ϫ 9 Յ 0. Note this is

equivalent to g1x2 Յ 0 for g1x2 ϭ Ϫx2 ϩ 6x Ϫ 9. Since a 6 0, the graph of g will

open downward. The factored form is g1x2 ϭ Ϫ1x Ϫ 32 2, showing 3 is a zero and a

repeated root. Using the x-axis, we plot the point (3, 0) and visualize a parabola

opening downward through this point.

Figure 3.29 shows the graph is below the x-axis (outputs are negative) for all

values of x except x ϭ 3. But since this is a less than or equal to inequality, the

solution is x ʦ ‫ޒ‬.

Since x ϭ 3 was a zero of

multiplicity 2, the graph “bounced

off” the x-axis at this point, with no

change of sign for g. The graph is

entirely below the x-axis, except at

the vertex (3, 0).

Graphical Solution

Figure 3.29

Ϫ1

0

1

2

3

4

5

6

7

x

aϽ0

The complete graph of g shown in Figure 3.30 confirms the analytical solution

(using the zeroes method). For the intervals of the domain shown in red:

1Ϫq, 32 ´ 13, q 2 , the graph of g is below the x-axis 3g1x2 6 0 4 . The point (3, 0)

is on the x-axis 3g132 ϭ 04 . As with the analytical solution, the solution to this

“less than or equal to” inequality is all real numbers. A calculator check of the

original inequality is shown in Figure 3.31.

Figure 3.30

Figure 3.31

y

10

2

Ϫ2

6

x

Ϫ2

8

g(x)

Ϫ3

Ϫ8

Now try Exercises 121 through 132

As an alternative to the Zeroes Method, the Interval Test Method can be used to

solve quadratic inequalities. Using the fact that all polynomials are continuous, test

values are selected from certain intervals of the domain and substituted into the original function.

Interval Test Method for Solving Inequalities

1. Find all real roots of the related equation (if they exist) and

plot them on the x-axis.

2. Select any convenient test value from each interval created

by the zeroes, and substitute these into the function.

3. The sign of the function at these test values will be the sign

of the function for all values of x in this interval.

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Section 3.2 Solving Quadratic Equations and Inequalities

EXAMPLE 9

Solve the inequality Ϫ2x2 Ϫ 3x ϩ 20 Յ 0 using interval tests.

Solution

To begin, we find the zeroes of f 1x2 ϭ Ϫ2x2 Ϫ 3x ϩ 20 by factoring.

Ϫ2x2 Ϫ 3x ϩ 20 ϭ 0

2x2 ϩ 3x Ϫ 20 ϭ 0

12x Ϫ 52 1x ϩ 42 ϭ 0

2x Ϫ 5 ϭ 0 or x ϩ 4 ϭ 0

5

x ϭ or x ϭ Ϫ4

2

related equation

multiply by ؊1

factored form

zero factor property

solutions

Plotting these intercepts creates three intervals on the x-axis (Figure 3.32).

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

Figure 3.32

1

0

1

2

3

4

5

6

2

7

8

7

8

3

Selecting a test value from each interval (in red) gives Figure 3.33:

Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

Figure 3.33

1

f 1Ϫ52 ϭ Ϫ15

f 1x2 6 0 in 1

When evaluating a function using the

interval test method, it’s usually

easier to use the factored form

instead of the polynomial form, since

all you really need is whether the

result will be positive or negative. For

instance, you could likely tell

f 1Ϫ52 ϭ Ϫ 321Ϫ52 Ϫ 5 4 1Ϫ5 ϩ 42 is

going to be negative, more

quickly than f 1Ϫ52 ϭ

Ϫ21Ϫ52 2 Ϫ 31Ϫ52 ϩ 20.

1

2

3

4

5

6

2

x ϭ Ϫ5

WORTHY OF NOTE

0

3

xϭ0

xϭ5

f 102 ϭ 20

f 152 ϭ Ϫ45

f 1x2 7 0 in 2

f 1x2 6 0 in 3

The interval tests show Ϫ2x2 Ϫ 3x ϩ 20 Յ 0

for x ʦ 1Ϫq, Ϫ42 ´ 1 52, q 2 , which is

supported by the graph shown in Figure 3.34.

Figure 3.34

30

Ϫ6

5

Ϫ15

Now try Exercises 133 through 144

The need to solve a quadratic inequality occurs in a variety of contexts. Here, the

solution is used to find the domain of a radical function.

EXAMPLE 10

Solving a Quadratic Inequality to Determine the Domain

Find the domain of g1x2 ϭ 29 Ϫ x2.

Solution

From our earlier work, the radicand must be nonnegative and we are essentially

asked to solve the inequality 9 Ϫ x2 7 0. The related equation is 9 Ϫ x2 ϭ 0 and

by inspection (or factoring), the solutions are x ϭ Ϫ3 and x ϭ 3. Plotting these

solutions creates three intervals on the x-axis (Figure 3.35).

Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

Figure 3.35

1

0

2

1

2

3

4

5

3

6

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Selecting a test value from each interval (in red) gives Figure 3.36:

Figure 3.36

Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1

1

x ϭ Ϫ4

9 Ϫ 1Ϫ42 ϭ Ϫ7

2

9 Ϫ x2 6 0 in 1

0

1

2

2

3

4

5

6

3

xϭ0

9 Ϫ 102 ϭ 9

2

9 Ϫ x2 7 0 in 2

The interval tests show that 9 Ϫ x2 Ն 0 for

x ʦ 3Ϫ3, 34 , which is the domain of

g1x2 ϭ 29 Ϫ x2. This is the same relation

we graphed in Section 1.1/Example 3, a

semicircle with endpoints at x ϭ Ϫ3 and 3.

See Figure 3.37.

xϭ4

9 Ϫ 142 2 ϭ Ϫ7

9 Ϫ x2 6 0 in 3

Figure 3.37

4

Ϫ6.1

6.1

Ϫ4

E. You’ve just seen how

inequalities

Now try Exercises 145 through 154

F. Applications of Quadratic Functions and Inequalities

A projectile is any object that is thrown, shot, or projected upward with no sustaining

source of propulsion. The height of the projectile at time t is modeled by the equation

h ϭ Ϫ16t2 ϩ vt ϩ k, where h is the height of the object in feet, t is the elapsed time in

seconds, and v is the initial velocity in feet per second. The constant k represents the

initial height of the object above ground level, as when a person releases an object 5 ft

above the ground in a throwing motion 1k ϭ 52, or when a rocket runs out of fuel at an

altitude of 240 ft 1k ϭ 2402 .

EXAMPLE 11

Solving an Application of Quadratic Equations — Rocketry

Figure 3.38

A model rocketry club is testing a newly developed engine. A few

seconds after liftoff, at a velocity of 160 ft/sec and a height of 240 ft,

it runs out of fuel and becomes a projectile (see Figure 3.38).

a. How high is the rocket 3 sec later?

b. For how many seconds was the height of the rocket greater

than or equal to 496 ft?

Projectile

phase

c. How many seconds until the rocket returns to the ground?

Solution

a. Using the information given, the function h modeling

the rocket’s height in the projectile phase, is

h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240. For its height at t ϭ 3 we

have

h132 ϭ Ϫ16132 2 ϩ 160132 ϩ 240

ϭ Ϫ16192 ϩ 480 ϩ 240

ϭ 576

Three seconds later, the rocket was at an altitude of 576 ft.

240 ft

Power

phase

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b. Since the height h(t) must be greater than or equal to 496 ft, we use the

function h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240 to write the inequality Ϫ16t2 ϩ 160t ϩ

240 Ն 496. In standard form, we obtain Ϫ16t2 ϩ 160t Ϫ 256 Ն 0 (subtract

496 from both sides). We begin by finding the zeroes of Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0,

noting the related graph opens downward since the leading coefficient is

negative.

Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0

t2 Ϫ 10t ϩ 16 ϭ 0

1t Ϫ 22 1t Ϫ 82 ϭ 0

t Ϫ 2 ϭ 0 or t Ϫ 8 ϭ 0

t ϭ 2 or t ϭ 8

related equation

divide by ؊16

factor

zero factor theorem

result

This shows the rocket is at exactly 496 ft after 2 sec (on its ascent) and after

8 sec (during its descent). We conclude the rocket’s height was greater than

496 ft for 8 Ϫ 2 ϭ 6 sec.

c. When the rocket hits the ground, its height is h ϭ 0. Substituting 0 for h1t2 and

solving gives

h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240

0 ϭ Ϫ16t2 ϩ 160t ϩ 240

0 ϭ t2 Ϫ 10t Ϫ 15

original function

substitute 0 for h(t )

divide by ؊16

The equation is nonfactorable, so we use the quadratic equation to solve, with

a ϭ 1, b ϭ Ϫ10, and c ϭ Ϫ15:

ϭ

Ϫb Ϯ 2b2 Ϫ 4ac

2a

Ϫ1Ϫ102 Ϯ 21Ϫ102 2 Ϫ 41121Ϫ152

2112

10 Ϯ 2160

2

10

4 210

ϭ

Ϯ

2

2

ϭ

substitute 1 for a, ؊10 for b, ؊15 for c

simplify

2160 ϭ 4 210

ϭ 5 Ϯ 2 210

simplify

Since we need the time t in seconds, we

use the approximate form of the answer,

obtaining t Ϸ Ϫ1.32 and t Ϸ 11.32. The

11 sec (since t represents time, the solution

t ϭ Ϫ1.32 does not apply). A calculator

check is shown in Figure 3.39 using the

Zeroes Method.

Figure 3.39

800

0

15

Ϫ200

Now try Exercises 157 through 166

EXAMPLE 12

Solving Applications of Inequalities Using the Quadratic Formula

For the years 1995 to 2006, the amount A of annual international telephone traffic

(in billions of minutes) can be modeled by A ϭ 0.17x2 ϩ 8.43x ϩ 64.58 where

x ϭ 0 represents the year 1995 [Source: Data from the 2009 Statistical Abstract of

the United States, Table 1344, page 846]. If this trend continues, in what year will

the annual number of minutes reach or surpass 275 billion?

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D. The Quadratic Formula and the Discriminant

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