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Non-Linear Inequalities
Joseph Lee
Metropolitan Community College
Joseph Lee Non-Linear Inequalities
Example 1.a
Solve the non-linear equation x2 − 7x + 12 = 0.
Solution: We know how to solve this equation:
x2 − 7x + 12 = 0
(x − 3)(x − 4) = 0
This left side of the equation will equal zero if x − 3 = 0 or ifx − 4 = 0. Thus, we have solutions of x = 3 or x = 4.
Joseph Lee Non-Linear Inequalities
Example 1.a
Solve the non-linear equation x2 − 7x + 12 = 0.
Solution: We know how to solve this equation:
x2 − 7x + 12 = 0
(x − 3)(x − 4) = 0
This left side of the equation will equal zero if x − 3 = 0 or ifx − 4 = 0. Thus, we have solutions of x = 3 or x = 4.
Joseph Lee Non-Linear Inequalities
Example 1.b
Solve the non-linear inequality x2 − 7x + 12 > 0.
Solution: Notice the similarity to the equation x2 − 7x − 12 = 0.Let’s proceed as before:
x2 − 7x + 12 > 0
(x − 3)(x − 4) > 0
We are looking for the left side of the equation to be positive. Weknow values of x = 3 and x = 4 will make the left side equal tozero. These values are called boundary points. To determinewhat values satisfy our inequality, we must look at all the regionsseparated by the boundary points. Thus, there are three differentcases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 1.b
Solve the non-linear inequality x2 − 7x + 12 > 0.
Solution: Notice the similarity to the equation x2 − 7x − 12 = 0.Let’s proceed as before:
x2 − 7x + 12 > 0
(x − 3)(x − 4) > 0
We are looking for the left side of the equation to be positive. Weknow values of x = 3 and x = 4 will make the left side equal tozero. These values are called boundary points. To determinewhat values satisfy our inequality, we must look at all the regionsseparated by the boundary points. Thus, there are three differentcases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 1.b (continued)
Case (1): If x is in the interval (−∞, 3). Then:◦ (x − 3) is negative,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (−∞, 3) is a solution.
Case (2): If x is in the interval (3, 4). Then:◦ (x − 3) is positive,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is negative.
Thus, any x in the interval (3, 4) is not a solution.Case (3): If x is in the interval (4,∞). Then:◦ (x − 3) is positive,◦ (x − 4) is positive,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (4,∞) is a solution.
Keep in mind that x = 3 and x = 4 are not solutions to theinequality. Therefore, we may conclude our solution is(−∞, 3) ∪ (4,∞).
Joseph Lee Non-Linear Inequalities
Example 1.b (continued)
Case (1): If x is in the interval (−∞, 3). Then:◦ (x − 3) is negative,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (−∞, 3) is a solution.Case (2): If x is in the interval (3, 4). Then:◦ (x − 3) is positive,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is negative.
Thus, any x in the interval (3, 4) is not a solution.
Case (3): If x is in the interval (4,∞). Then:◦ (x − 3) is positive,◦ (x − 4) is positive,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (4,∞) is a solution.
Keep in mind that x = 3 and x = 4 are not solutions to theinequality. Therefore, we may conclude our solution is(−∞, 3) ∪ (4,∞).
Joseph Lee Non-Linear Inequalities
Example 1.b (continued)
Case (1): If x is in the interval (−∞, 3). Then:◦ (x − 3) is negative,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (−∞, 3) is a solution.Case (2): If x is in the interval (3, 4). Then:◦ (x − 3) is positive,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is negative.
Thus, any x in the interval (3, 4) is not a solution.Case (3): If x is in the interval (4,∞). Then:◦ (x − 3) is positive,◦ (x − 4) is positive,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (4,∞) is a solution.
Keep in mind that x = 3 and x = 4 are not solutions to theinequality. Therefore, we may conclude our solution is(−∞, 3) ∪ (4,∞).
Joseph Lee Non-Linear Inequalities
Example 1.b (continued)
Case (1): If x is in the interval (−∞, 3). Then:◦ (x − 3) is negative,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (−∞, 3) is a solution.Case (2): If x is in the interval (3, 4). Then:◦ (x − 3) is positive,◦ (x − 4) is negative,◦ so (x − 3)(x − 4) is negative.
Thus, any x in the interval (3, 4) is not a solution.Case (3): If x is in the interval (4,∞). Then:◦ (x − 3) is positive,◦ (x − 4) is positive,◦ so (x − 3)(x − 4) is positive.
Thus, any x in the interval (4,∞) is a solution.
Keep in mind that x = 3 and x = 4 are not solutions to theinequality. Therefore, we may conclude our solution is(−∞, 3) ∪ (4,∞).
Joseph Lee Non-Linear Inequalities
Example 2.a
Solve the non-linear inequality x2 + x ≤ 20.
Solution: We want to make one side of this inequality zero.
x2 + x ≤ 20
x2 + x − 20 ≤ 0
(x + 5)(x − 4) ≤ 0
We are looking for the left side of the equation to be negative orequal to zero. We know values of x = −5 and x = 4 will make theleft side equal to zero – these are our boundary points. Thus, thereare three different cases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 2.a
Solve the non-linear inequality x2 + x ≤ 20.
Solution: We want to make one side of this inequality zero.
x2 + x ≤ 20
x2 + x − 20 ≤ 0
(x + 5)(x − 4) ≤ 0
We are looking for the left side of the equation to be negative orequal to zero. We know values of x = −5 and x = 4 will make theleft side equal to zero – these are our boundary points. Thus, thereare three different cases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 2.a (continued)
Case (1): If x is in the interval (−∞,−5). Then:◦ (x + 5) is negative,◦ (x − 4) is negative,◦ so (x + 5)(x − 4) is positive.
Thus, any x in the interval (−∞,−5) is not a solution.Case (2): If x is in the interval (−5, 4). Then:◦ (x + 5) is positive,◦ (x − 4) is negative,◦ so (x + 5)(x − 4) is negative.
Thus, any x in the interval (−5, 4) is a solution.Case (3): If x is in the interval (4,∞). Then:◦ (x + 5) is positive,◦ (x − 4) is positive,◦ so (x + 5)(x − 4) is positive.
Thus, any x in the interval (4,∞) is not a solution.
Keep in mind that x = −5 and x = 4 are solutions to theinequality. Therefore, we may conclude our solution is [−5, 4].
Joseph Lee Non-Linear Inequalities
Example 2.b
Solve the non-linear inequality 4x2 ≥ 4x + 3.
Solution: We want to make one side of this inequality zero.
4x2 ≥ 4x + 3
4x2 − 4x − 3 ≥ 0
(2x + 1)(2x − 3) ≥ 0
We are looking for the left side of the equation to be positive orequal to zero. We know values of x = −1
2 and x = 32 will make the
left side equal to zero – these are our boundary points. Thus, thereare three different cases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 2.b
Solve the non-linear inequality 4x2 ≥ 4x + 3.
Solution: We want to make one side of this inequality zero.
4x2 ≥ 4x + 3
4x2 − 4x − 3 ≥ 0
(2x + 1)(2x − 3) ≥ 0
We are looking for the left side of the equation to be positive orequal to zero. We know values of x = −1
2 and x = 32 will make the
left side equal to zero – these are our boundary points. Thus, thereare three different cases we need to consider:
Joseph Lee Non-Linear Inequalities
Example 2.b (continued)
Case (1): If x is in the interval(−∞,−1
2
). Then:
◦ (2x + 1) is negative,◦ (2x − 3) is negative,◦ so (2x + 1)(2x − 3) is positive.
Thus, any x in the interval(−∞,−1
2
)is a solution.
Case (2): If x is in the interval(−1
2 ,32
). Then:
◦ (2x + 1) is positive,◦ (2x − 3) is negative,◦ so (2x + 1)(2x − 3) is negative.
Thus, any x in the interval(−1
2 ,32
)is not a solution.
Case (3): If x is in the interval(32 ,∞
). Then:
◦ (2x + 1) is positive,◦ (2x − 3) is positive,◦ so (2x + 1)(2x − 3) is positive.
Thus, any x in the interval(32 ,∞
)is a solution.
Keep in mind that x = −12 and x = 3
2 are solutions to theinequality. Therefore, we may conclude our solution is(−∞,−1
2
]∪[32 ,∞
).
Joseph Lee Non-Linear Inequalities
Example 3.a
Solve the non-linear inequalityx − 3
x + 4≥ 0.
Solution: With a rational inequality, we need to examine both thenumerator and the denominator. The value x = 3 makes thenumerator equal to 0, and if the numerator is 0, then the value ofthe expression is 0. The value x = −4 makes the denominatorequal to 0, and if the denominator is 0, then the expression isundefined. The boundary points for rational inequalities includeany value that makes numerator or denominator equal to 0. Wenow consider three different cases:
Joseph Lee Non-Linear Inequalities
Example 3.a
Solve the non-linear inequalityx − 3
x + 4≥ 0.
Solution: With a rational inequality, we need to examine both thenumerator and the denominator. The value x = 3 makes thenumerator equal to 0, and if the numerator is 0, then the value ofthe expression is 0. The value x = −4 makes the denominatorequal to 0, and if the denominator is 0, then the expression isundefined. The boundary points for rational inequalities includeany value that makes numerator or denominator equal to 0. Wenow consider three different cases:
Joseph Lee Non-Linear Inequalities
Example 3.a (continued)
Case (1): If x is in the interval (−∞,−4). Then:◦ (x − 3) is negative,◦ (x + 4) is negative,
◦ sox − 3
x + 4is positive.
Thus, any x in the interval (−∞,−4) is a solution.
Case (2): If x is in the interval (−4, 3). Then:◦ (x − 3) is negative,◦ (x + 4) is positive,
◦ sox − 3
x + 4is negative.
Thus, any x in the interval (−4, 3) is not a solution.
Case (3): If x is in the interval (3,∞). Then:◦ (x − 3) is positive,◦ (x + 4) is positive,
◦ sox − 3
x + 4is positive.
Thus, any x in the interval (3,∞) is a solution.
Joseph Lee Non-Linear Inequalities
Example 3.a (continued)
Keep in mind that x = 3 is a solution to the inequality, butx = −4 is not a solution to the inequality (as it not in the domainof the expression). Therefore, we may conclude our solution is(−∞,−4) ∪ [3,∞).
Joseph Lee Non-Linear Inequalities
Example 3.b
Solve the non-linear inequality2x2 − 5x + 3
2− x≥ 0.
Solution:2x2 − 5x + 3
2− x≥ 0
(2x − 3)(x − 1)
2− x≥ 0
Our boundary points are any values that make the expression equalto zero or undefined. The values are x = 3
2 and x = 1 make theexpression equal to zero, and the value x = 2 make the expressionundefined. We now consider four different cases:
Joseph Lee Non-Linear Inequalities
Example 3.b
Solve the non-linear inequality2x2 − 5x + 3
2− x≥ 0.
Solution:2x2 − 5x + 3
2− x≥ 0
(2x − 3)(x − 1)
2− x≥ 0
Our boundary points are any values that make the expression equalto zero or undefined. The values are x = 3
2 and x = 1 make theexpression equal to zero, and the value x = 2 make the expressionundefined. We now consider four different cases:
Joseph Lee Non-Linear Inequalities
Example 3.b (continued)
Case (1): If x is in the interval (−∞, 1). Then:
◦ (2x − 3) is negative,◦ (x − 1) is negative,◦ (2− x) is positive,
◦ so(2x − 3)(x − 1)
2− xis positive.
Thus, any x in the interval (−∞, 1) is not a solution.
Case (2): If x is in the interval(1, 32
). Then:
◦ (2x − 3) is negative,◦ (x − 1) is positive,◦ (2− x) is positive,
◦ so(2x − 3)(x − 1)
2− xis negative.
Thus, any x in the interval(1, 32
)is a solution.
Joseph Lee Non-Linear Inequalities
Example 3.b (continued)
Case (3): If x is in the interval(32 , 2
). Then:
◦ (2x − 3) is positive,◦ (x − 1) is positive,◦ (2− x) is positive,
◦ so(2x − 3)(x − 1)
2− xis positive.
Thus, any x in the interval(32 , 2
)is not a solution.
Case (4): If x is in the interval (2,∞). Then:◦ (2x − 3) is positive,◦ (x − 1) is positive,◦ (2− x) is negative,
◦ so(2x − 3)(x − 1)
2− xis negative.
Thus, any x in the interval (2,∞) is a solution.
Keep in mind that x = 1 and x = 32 are solutions to the inequality,
but x = 2 is not a solution to the inequality (as it not in thedomain of the expression). Therefore, we may conclude oursolution is
[1, 32
]∪ (2,∞).
Joseph Lee Non-Linear Inequalities
Example 3.b (continued)
Case (3): If x is in the interval(32 , 2
). Then:
◦ (2x − 3) is positive,◦ (x − 1) is positive,◦ (2− x) is positive,
◦ so(2x − 3)(x − 1)
2− xis positive.
Thus, any x in the interval(32 , 2
)is not a solution.
Case (4): If x is in the interval (2,∞). Then:◦ (2x − 3) is positive,◦ (x − 1) is positive,◦ (2− x) is negative,
◦ so(2x − 3)(x − 1)
2− xis negative.
Thus, any x in the interval (2,∞) is a solution.
Keep in mind that x = 1 and x = 32 are solutions to the inequality,
but x = 2 is not a solution to the inequality (as it not in thedomain of the expression). Therefore, we may conclude oursolution is
[1, 32
]∪ (2,∞).
Joseph Lee Non-Linear Inequalities
Example 4
Solve the non-linear inequalityx
x + 4≥ 2.
Solution:x
x + 4≥ 2
x
x + 4− 2 ≥ 0
x
x + 4− 2 · x + 4
x + 4≥ 0
x
x + 4− 2x + 8
x + 4≥ 0
x − (2x + 8)
x + 4≥ 0
Joseph Lee Non-Linear Inequalities
Example 4
Solve the non-linear inequalityx
x + 4≥ 2.
Solution:x
x + 4≥ 2
x
x + 4− 2 ≥ 0
x
x + 4− 2 · x + 4
x + 4≥ 0
x
x + 4− 2x + 8
x + 4≥ 0
x − (2x + 8)
x + 4≥ 0
Joseph Lee Non-Linear Inequalities
Example 4 (continued)
x − (2x + 8)
x + 4≥ 0
x − 2x − 8
x + 4≥ 0
−x − 8
x + 4≥ 0
Our boundary points are any values that make the expression equalto zero or undefined. The value are x = −8 make the expressionequal to zero, and the value x = −4 make the expressionundefined. We now consider three different cases:
Joseph Lee Non-Linear Inequalities
Example 4 (continued)
Case (1): If x is in the interval (−∞,−8). Then:◦ (−x − 8) is positive,◦ (x + 4) is negative,
◦ so−x − 8
x + 4is negative.
Thus, any x in the interval (−∞,−8) is not a solution.
Case (2): If x is in the interval (−8,−4). Then:◦ (−x − 8) is negative,◦ (x + 4) is negative,
◦ so−x − 8
x + 4is positive.
Thus, any x in the interval (−8,−4) is a solution.
Case (3): If x is in the interval (−4,∞). Then:◦ (−x − 8) is negative,◦ (x + 4) is positive,
◦ so−x − 8
x + 4is negative.
Thus, any x in the interval (−4,∞) is not a solution.
Joseph Lee Non-Linear Inequalities
Example 4 (continued)
Since x = −8 makes the expression equal to zero, it is a solution.However, since x = −4 makes the expression undefined, it is not asolution. Thus, the solution is [−8,−4).
Joseph Lee Non-Linear Inequalities
