Beginning Algebra, 11th Edition

The fact that the graph of a quadratic function can intersect the x-axis in two, one, or

no points justifies why some quadratic equations have two, some have one, and some

have no real solutions.

584 CHAPTER 9 Quadratic Equations

x-Intercepts of the Graph of a Quadratic Function

The real number solutions of a quadratic equation are the

x-values of the x-intercepts of the graph of the corresponding quadratic function

defined by ƒ 1 x 2 =ax^2 +bx+c.

ax^2 +bx+c= 0

NOW TRY
EXERCISE 3
Decide from the graph how
many real number solutions
there are of the corresponding
equation Give the
solution set.

ƒ 1 x 2 =0.

0 x

–4

2

y

f(x) = –x^2 + 4x – 4

NOW TRY ANSWER
3.one real solution; 526

x

y

f(x) = x^2 – 3

2

0

–3

–4 –2 2 4

4

FIGURE 5

x

y

f(x) = x^2 – 4x + 4

–1^01

1

4

9

2345

FIGURE 6

(b)FIGURE 6shows the graph of The corresponding equation,

has one real solution, 2, which is the x-value of the x-intercept of the graph. The

solution set is

(c) FIGURE 7shows the graph of The

equation

has no real solutions, since there are no x-intercepts.

The solution set over the domain of real numbers is

The equation doeshave two pure imaginary

solutions: and i 22 - i 22 .B

0. A

x^2 + 2 = 0

ƒ 1 x 2 =x^2 +2.

526.

x^2 - 4 x+ 4 =0,

ƒ 1 x 2 =x^2 - 4 x+4.

x

y

f(x) = x^2 + 2

6

2

–4 –2^024

FIGURE 7

NOW TRY

Determining the Number of Real Solutions from Graphs

Decide from the graphs in FIGURES 5 –7how many real number solutions there are of

the corresponding equation. Give the solution set for the domain of real

numbers.

(a)FIGURE 5shows the graph of The corresponding equation,

,

has two real solutions, and which correspond to the x-intercepts. The

solution set is E (^23) F.

23 - 23 ,

x^2 - 3 = 0

ƒ 1 x 2 =x^2 - 3.

ƒ 1 x 2 = 0

EXAMPLE 3

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