A. Zeroes of Quadratic Functions and x-Intercepts of Quadratic Graphs
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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.
Quadratic Functions
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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Section 3.2 Solving Quadratic Equations and Inequalities
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
quadratic function and the
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
xϭ
or x ϭ Ϫ
A4
A4
3
3
xϭ
or x ϭ Ϫ
2
2
original equation
subtract 3, divide by ؊4
square root property of equality
simplify radicals
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
simplify radicals
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.