Transcript 1.6 Quadratic and Rational Inequalities

1.6 Quadratic and Rational
Inequalities
A quadratic inequality is any
inequality that can be put in
one of the forms
ax  bx  c  0 ax  bx  c  0
2
2
ax  bx  c  0 ax  bx  c  0
where a, b, and c are real numbers
and a  0.
2
2
Solving Quadratic Inequalities
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Step 1: write the inequality in standard form.
Step 2: solve the related quadratic equation.
Step 3: locate the boundary points on a
number line.
Step 4: construct a sign chart.
Step 5: the solution set is the interval that
produced a true statement.
Example 1: Solve a Quadratic
Inequality
Solve: x  2 x  3.
2
Solution:
Step 1: Write the inequality in
standard form.
x  2x  3  0
2
Step 2: Solve the related quadratic equation.
x2  2 x  3  0
x2  2 x  3  0
( x  1)( x  3)  0
x  1  0 or x  3  0
x  1 or x  3
The boundary points are -1 and 3.
1
3
These two points divide the number
line into three test intervals, namely
(-,-1), (-1,3), and (3,).
Take a test point within each
interval and check the sign.
Test
Interval
(, 1)
(1,3)
(3, )
Test
Point
X+1
X-3
(x+1)(x-3)
2


0


0
4


0
0
The question is "where x  2x  3  0?"
2
Our table shows that
x  2 x  3  ( x  1)( x  3)  0
in the interval (1,3).
2
So the solution set is
the interval (1,3).
Practice Exercises
Answers:
Solve:
1. x  4 x  3  0
2
2. x  2 x  1  0
2
3. 3 x  16 x  5
2
(1,3)
All reals
[5,  13 ]
Solving Rational Inequalities
x5
Solve:
0
x2
It is incorrect to multiply both sides by
x  2 to clear fractions. The problem is
that x  2 contains a variable and can be
positive or negative, depending on the
value of x. Thus, we do not know
whether or not to reverse the sense of
inequality.
x5
Example 2: Solve: x  2  0
Solution: We begin by finding values
of x that make the numerator and
denominator 0.
Set the numerator and denominator equal to 0.
x5  0
Solve
x  5
x2  0
x2
The boundary points are -5 and 2.
Locate boundary points -5 and 2
on a number line.
5
2
These boundary points divide the
number line into three intervals,
namely (, 5), (5, 2), and (2, ).
Now, construct a sign chart: take
one test point from each interval
and check the signs.
5
2
Intervals Test X+5
Points
X-2
x5
x2


0
0


0
3


0
(, 5)
6
(5, 2)
(2, )
The given question is "for what values
x5
of x the quantity
is positive?"
x2
x5
Our table shows that
 0 for
x2
(, 5) or (2, ).
Thus the solution set is
(, 5) or (2, ).
The graph of the solution set on a
number line is shown as follows:

)
5
(
2

Practice Exercises
Solve:
x5
1.
0
x2
1
2.
1
x3
x
3.
2
x2
Answers:
(5, 2)
(,3) or [4, )
[4, 2)