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Section 3.2 Solving Quadratic Equations and Inequalities

As shown, the equation is in standard form, meaning the terms are in decreasing

order of degree and the equation is set equal to zero. The family of quadratic functions

is similarly defined.

A quadratic function is one that can be written in the form

f 1x2 ϭ ax2 ϩ bx ϩ c,

where a, b, and c are real numbers and a

0.

In Section R.4, we noted that some quadratic equations have two real solutions, others

have only one, and still others have none. When these possibilities are explored graphically, we note a clear connection between the zeroes of a quadratic function and the

x-intercepts of its graph.

EXAMPLE 1

Solution

Noting Relationships between Zeroes and x-Intercepts

Consider the functions f 1x2 ϭ x2 Ϫ 2x Ϫ 3 and g1x2 ϭ x2 Ϫ 4x ϩ 4.

a. Find the zeroes of each function algebraically.

b. Find the x-intercepts of each function graphically.

c. Comment on how the zeroes and x-intercepts are related.

a. To find the zeroes algebraically, replace f (x) and g(x) with 0, then solve. In

each case, the solutions can be found by factoring.

For f (x):

x2 Ϫ 2x Ϫ 3 ϭ 0

1x Ϫ 321x ϩ 12 ϭ 0

x ϭ 3 or x ϭ Ϫ1

For g(x):

x2 Ϫ 4x ϩ 4 ϭ 0

1x Ϫ 221x Ϫ 22 ϭ 0

x ϭ 2 or x ϭ 2

There are two real solutions,

There is only one real solution,

x ϭ 3 and x ϭ Ϫ1.

but it is repeated twice.

b. To find the x-intercepts, go to the Y= screen and enter the first function as Y1

with Y2 ϭ 0 (the x-axis), then graph them in the standard window. Locate the

x-intercepts using the 2nd TRACE (CALC) 5:Intersect feature.

• For f 1x2 ϭ x2 Ϫ 2x Ϫ 3

Figure 3.9

Figure 3.10

10

Ϫ10

10

10

Ϫ10

Ϫ10

10

Ϫ10

The result shows there are two x-intercepts, which occur at x ϭ Ϫ1 (Figure 3.9)

and x ϭ 3 (Figure 3.10). These were also the real zeroes of f 1x2 ϭ x2 Ϫ 2x Ϫ 3.

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• For g1x2 ϭ x2 Ϫ 4x ϩ 4

After entering x2 Ϫ 4x ϩ 4 as Y1 (leaving Y2 ϭ 0) we again graph both functions

in the standard window. This time we find there is only one x-intercept, located at

x ϭ 2 (Figure 3.11), a fact supported by the accompanying table (Figure 3.12).

The graph is tangent to the x-axis at x ϭ 2 (touching the axis at just this one

point) and a closer look at g reveals why — the function is easily rewritten as

g1x2 ϭ 1x Ϫ 22 2, producing the zero at x ϭ 2 and positive values for all other

inputs! Note that x ϭ 2 was the only zero found for g1x2 ϭ x2 Ϫ 4x ϩ 4.

Figure 3.11

Figure 3.12

3

Ϫ5

5

Ϫ3

c. From parts (a) and (b), it’s apparent that the real zeroes of a function appear

graphically as x-intercepts.

Now try Exercises 7 through 20

In Section 3.1, we noted that the equation

x2 ϩ 1 ϭ 0 had no real solutions, indicating

the function f 1x2 ϭ x2 ϩ 1 has no real zeroes.

Here we might wonder whether a graphical

connection also exists for the “no real zeroes”

case. The graph of f 1x2 ϭ x2 ϩ 1 is given

in Figure 3.13 and shows that f has no

x-intercepts, further affirming the graphical

connection noted in Example 1. A summary of

these connections is shown in Figure 3.14 for

the case where a 7 0 (the graph opens upward). Similar statements can be made when

a 6 0 (the graph opens downward).

Figure 3.13

10

Ϫ10

10

Ϫ10

Figure 3.14

y

y

7

y

7

Ϫ3

7

Ϫ3

x

7

Ϫ3

7

Ϫ3

f 1x2 ϭ ax ϩ bx ϩ c

has two x-intercepts

and two real zeroes;

ax2 ϩ bx ϩ c ϭ 0

has two real solutions.

2

x

Ϫ3

7

Ϫ3

g1x2 ϭ ax ϩ bx ϩ c

has one x-intercept

and one real repeated zero;

ax2 ϩ bx ϩ c ϭ 0

has one real repeated solution.

2

h1x2 ϭ ax2 ϩ bx ϩ c

has no x-intercepts

and no real zeroes;

ax2 ϩ bx ϩ c ϭ 0

has no real solutions.

x

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Also from our work in Example 1, we learn that the following statements are equivalent, meaning that if any one of the statements is true, then all four statements are true.

A. You’ve just seen how

we can establish a relationship

between the zeroes of a

x-intercepts of its graph

For any real number r,

1. x ϭ r is a solution of f 1x2 ϭ 0.

2. r is a zero of f 1x2 .

3. 1x Ϫ r2 is a factor of f 1x2 .

4. (r, 0) is an x-intercept of the graph of f.

B. Quadratic Equations and the Square Root Property of Equality

In Section 1.5 we solved the equation ax ϩ b ϭ c for x to establish a general solution

for equations of this form. In this section, we’ll establish a general solution for

ax2 ϩ bx ϩ c ϭ 0 using a process known as completing the square. To begin, we note

that the equation x2 ϭ 9 can be solved by factoring. In standard form we have

x2 Ϫ 9 ϭ 0 (note b ϭ 02, then 1x ϩ 32 1x Ϫ 32 ϭ 0. The solutions are x ϭ Ϫ3 or

x ϭ 3, which are simply the positive and negative square roots of 9. This result suggests an alternative method for solving equations of the form X2 ϭ k, known as the

square root property of equality.

Square Root Property of Equality

If X represents an algebraic expression and X2 ϭ k,

then X ϭ 1k or X ϭ Ϫ 1k;

also written as X ϭ Ϯ 1k

EXAMPLE 2

Solving an Equation Using the Square Root Property of Equality

Use the square root property of equality to solve each equation. Verify solutions

graphically.

a. Ϫ4x2 ϩ 3 ϭ Ϫ6

b. x2 ϩ 12 ϭ 0

c. 1x Ϫ 52 2 ϭ 7

Solution

WORTHY OF NOTE

In Section R.6 we noted that for

any real number a, 2a2 ϭ ͿaͿ. From

Example 2(a), solving the equation

by taking the square root of both

sides produces 2x2 ϭ 294 . This is

equivalent to ͿxͿ ϭ 294 , again

showing this equation must have

two solutions, x ϭ Ϫ 294 and

x ϭ 294 .

a. Ϫ4x2 ϩ 3 ϭ Ϫ6

9

x2 ϭ

4

9

9

or x ϭ Ϫ

A4

A4

3

3

or x ϭ Ϫ

2

2

original equation

subtract 3, divide by ؊4

square root property of equality

This equation has two rational solutions.

Using the Intersection Method with Y1 ϭ Ϫ4x2 ϩ 3 and Y2 ϭ Ϫ6, we note

the graphs intersect at two points. This shows there will be two real roots,

which turn out to be rational (Figures 3.15 and 3.16).

Figure 3.15

Figure 3.16

5

Ϫ5

5

5

Ϫ10

Ϫ5

5

Ϫ10

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b. x2 ϩ 12 ϭ 0

x2 ϭ Ϫ12

x ϭ 1Ϫ12 or x ϭ Ϫ 1Ϫ12

x ϭ 2i 13 or x ϭ Ϫ2i 13

original equation

subtract 12

square root property of equality

This equation has two complex solutions.

Using the Zeroes Method with Y1 ϭ x2 ϩ 12 shows there are no x-intercepts,

and therefore no real roots (Figure 3.17). However, the roots can still be

checked on the home screen, as shown in Figure 3.18.

Figure 3.17

Figure 3.18

30

Ϫ10

10

Ϫ10

c. 1x Ϫ 52 2 ϭ 7

x Ϫ 5 ϭ 17

x ϭ 5 ϩ 27

original equation

or

x Ϫ 5 ϭ Ϫ 17

x ϭ 5 Ϫ 27

square root property of equality

solve for x

This equation has two irrational solutions.

Using the Zeroes Method with Y1 ϭ 1x Ϫ 52 2 Ϫ 7, we note the graph has two

x-intercepts. This shows there will be two real roots, which turn out to be

irrational (Figures 3.19 and 3.20).

Figure 3.19

Figure 3.20

10

10

Ϫ2

10

Ϫ10

Ϫ2

10

Ϫ10

Now try Exercises 21 through 36

CAUTION

For equations of the form 1x ϩ d2 2 ϭ k as in Example 2(c), you should resist the temptation to expand the binomial square in an attempt to simplify the equation and solve by

factoring—many times the result is nonfactorable. Any equation of the form 1x ϩ d2 2 ϭ k

can quickly be solved using the square root property of equality.