Download Solving Nonlinear Inequalities

Digital Lesson
Solving Nonlinear
Inequalities
A quadratic inequality in one variable is an inequality which
can be written in the form
ax2 + bx + c > 0 (a  0)
for a, b, c real numbers.
The symbols , , and  may also be used.
Example: x2 – 3x + 7  0 is a quadratic inequality since it can
be written 1x2 + (–3)x + 7  0.
Example: 3x2 < x + 5 is a quadratic inequality since it can be
written 3x2 + (–1)x + (–5) < 0.
Example: x2 + 3x  x2 + 4 is not a quadratic inequality since it is
equivalent to 3x  4  0.
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2
A solution of a quadratic inequality in one variable is a number
which, when substituted for the variable, results in a true
inequality.
Example: Which of the values of x are solutions of x2 + 3x  4  0 ?
x2 + 3x – 4  0
x
x2 + 3x – 4
1
(1)2 + 3(1) – 4
 6  0 true
yes
(0)2 + 3(0) – 4
 4  0 true
yes
 2.25  0 true
yes
0
0.5
(0.5)2 + 3(0.5) – 4
Solution?
1
(1)2 + 3(1) – 4
0  0 true
yes
2
(2)2 + 3(2) – 4
6  0 false
no
3
(3)2 + 3(3) – 4
14  0 false
no
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The solution set of an inequality is the set of all solutions.
Study the graph of the solution set of x2 + 3x  4  0.
[
-6 -5 - 4 -3 -2 -1
]
0
1
2
The solution set is {x |  4  x  1}.
The values of x for which equality holds are part of the
solution set.
These values can be found by solving the quadratic equation
associated with the inequality.
x2 + 3x  4 = 0 Solve the associated equation.
(x + 4)(x  1) = 0 Factor the trinomial.
x =  4 or x = 1 Solutions of the equation
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4
To solve a quadratic inequality:
1. If necessary, rewrite the quadratic inequality so that zero
appears on the right, then factor.
2. On the real number line, draw a vertical line at the
numbers that make each factor equal to zero.
3. For each factor, place plus signs above the number line in
the regions where the factor is positive, and minus signs
where the factor is negative.
4. Observe the sign of the product of the factors for each region,
to determine which regions will belong to the solution set.
5. Express the solution set using set-builder notation and a
graph on a real number line.
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Example: Solve and graph the solution set of x2  6x + 5 < 0.
The product of the
factors is negative.
x–1
x–5
(x  1)(x  5) < 0 Factor.
x  1 = 0 x  5 = 0 Solve for each factor equal
x=1
x = 5 to zero.
Draw vertical lines indicating
Product is Product is Product is the numbers where each factor
positive.
negative. positive. equals zero.
––– +++++
––– –––––
(
Factors
-1
0
1
)
2
3 4
5
{x | 1 < x < 5}
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+++
+++
6
7
For each region, identify if each
factor is positive or negative.
Draw the solution set.
Rounded parentheses indicate a
strict inequality.
Solution set in set-builder
notation.
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Example: Solve and graph the solution set of x2  x  6.
x2  x  6  0
The product of the
factors is positive.
(x + 2)(x  3)  0
x =  2, 3
x+2 –––– +++++++ ++
x–3 –––– ––––––– ++
]
- 4 -3 -2 -1
[
0
1
2
3 4
Rewrite the inequality so that
zero appears on the right.
Factor.
Numbers where each
factor equals zero.
Draw vertical lines where each
factor equals zero.
Indicate positive and negative
regions for each factor.
Draw solution set.
Square brackets are used since
the inequality is .
{x | x   2 or x  3}
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Solution set in set-builder
notation.
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Cubic inequalities can be solved similarly.
Example: Solve and graph the solution set of x3 + x2  9x  9 > 0.
x2(x + 1)  9(x + 1) > 0
(x2  9)(x + 1) > 0
Factor by grouping.
(x + 3)(x  3)(x + 1) > 0
x =  3, +3, 1
x–3 ––
x+3 ––
x+1 ––
––
(
––––– ++
++ +++++ ++
–– +++++ ++
)
- 4 -3 -2 -1
(
0
1
2
Draw three vertical lines.
Indicate positive and negative
regions for each of the three
factors.
3 4
{x | 3 < x < 1 or x > 3}
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Numbers where each factor
equals zero.
Solution set
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Inequalities involving rational functions can be solved similarly.
( x  1)
 0.
Example: Solve and graph the solution set of
( x  2)
(x + 1) = 0 (x  2) = 0
Find the numbers for which
each factor equals zero.
x = 1
x=2
Note that 2 will not be part of the solution set since the
expression is not defined when the denominator is zero.
x+1 ––––– ++++ +++
x–2 ––––– –––– +++
]
- 4 -3 -2 -1
(
0
1
2
3 4
{x | x   1 or x > 2}
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There are two regions
where the quotient of the
two factors is positive.
Solution set
9
( x  2)
 0.
Example: Solve and graph the solution set of 2
( x  2 x  3)
The quotient
( x  2)
0
Factor.
is negative.
( x  1)( x  3)
x+2=0
x = 2
(x  1)(x + 3) = 0
x = 1, 3
Expression is undefined
at these points.
x+2 –– – ++++ +++++
x1 –– – –––– +++++
x+3 –– + ++++ +++++
) (
- 4 -3 -2 -1
)
0
1
2
3 4
{x | x < 3 or 2 < x < 1} Solution set
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10
Example: One leg of a right triangle is 2 inches longer than the
other. How long should the shorter leg be to ensure that
the area of the triangle is greater than or equal to 4?
x = shorter leg
x + 2 = other leg
x
4
1
Area of triangle  base  height
2
x+2
1
Solve: ( x  2)( x)  4
x+4 – ++++++++ +++
2
( x  2)( x )  8
x–2 – –––––––– +++
]
[
x2  2x  8  0
- 4 -3 -2 -1 0 1 2 3 4
( x  4)( x  2)  0
Since length has to be positive, the
x   4 or x  2
answer is x  2.
The shorter leg should be at least 2 inches long.
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