H Algebra 2/Trig · Unit 4
Solve rational inequalities with critical numbers, a sign check, and interval notation.
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One side zero
Move every term to one side so the other side is 0.
Write the left side as one fraction.
Critical numbers
Critical numbers come from the numerator and from the denominator.
They split the number line into sections.
Divide-by-zero
A denominator of 0 is never a solution.
Mark that point with an open circle.
Interval notation
A bracket includes a numerator zero when the symbol is ≤ or ≥.
A parenthesis does not include the endpoint. Use ∪ between separate sections.
One fraction compared with 0
The LCD is (x + 1).
Write one fraction.
x − 5(x + 1) ≤ 0
Factor the numerator.
(x − 5)(x + 1) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = −1, x = 5
Divide-by-zero values are never solutions.
x ≠ −1
These numerator zeros are solutions because the symbol includes equals.
x = 5
Try x = −2.
The value is 7, which is positive.
Try x = 2.
The value is −1, which is negative.
Try x = 6.
The value is 17, which is positive.
(−1, 5]
Answer: (−1, 5]
The LCD is (x − 4).
Write one fraction.
x + 3(x − 4) ≥ 0
Factor the numerator.
(x + 3)(x − 4) ≥ 0
Critical numbers come from the numerator and from the denominator.
x = −3, x = 4
Divide-by-zero values are never solutions.
x ≠ 4
These numerator zeros are solutions because the symbol includes equals.
x = −3
Try x = −4.
The value is 18, which is positive.
Try x = 12.
The value is −1, which is negative.
Try x = 5.
The value is 8, which is positive.
(−∞, −3] ∪ (4, ∞)
Answer: (−∞, −3] ∪ (4, ∞)
The LCD is (x + 6).
Write one fraction.
x − 2(x + 6) < 0
Factor the numerator.
(x − 2)(x + 6) < 0
Critical numbers come from the numerator and from the denominator.
x = −6, x = 2
Divide-by-zero values are never solutions.
x ≠ −6
Try x = −7.
The value is 9, which is positive.
Try x = −2.
The value is −1, which is negative.
Try x = 3.
The value is 19, which is positive.
(−6, 2)
Answer: (−6, 2)
The LCD is (x − 3).
Write one fraction.
x + 1(x − 3) > 0
Factor the numerator.
(x + 1)(x − 3) > 0
Critical numbers come from the numerator and from the denominator.
x = −1, x = 3
Divide-by-zero values are never solutions.
x ≠ 3
Try x = −2.
The value is 15, which is positive.
Try x = 1.
The value is −1, which is negative.
Try x = 4.
The value is 5, which is positive.
(−∞, −1) ∪ (3, ∞)
Answer: (−∞, −1) ∪ (3, ∞)
The LCD is (x − 1).
Write one fraction.
x − 8(x − 1) ≤ 0
Factor the numerator.
(x − 8)(x − 1) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = 1, x = 8
Divide-by-zero values are never solutions.
x ≠ 1
These numerator zeros are solutions because the symbol includes equals.
x = 8
Try x = 0.
The value is 8, which is positive.
Try x = 92.
The value is −1, which is negative.
Try x = 9.
The value is 18, which is positive.
(1, 8]
Answer: (1, 8]
The LCD is (x + 2).
Write one fraction.
x + 5(x + 2) ≥ 0
Factor the numerator.
(x + 5)(x + 2) ≥ 0
Critical numbers come from the numerator and from the denominator.
x = −5, x = −2
Divide-by-zero values are never solutions.
x ≠ −2
These numerator zeros are solutions because the symbol includes equals.
x = −5
Try x = −6.
The value is 14, which is positive.
Try x = −72.
The value is −1, which is negative.
Try x = −1.
The value is 4, which is positive.
(−∞, −5] ∪ (−2, ∞)
Answer: (−∞, −5] ∪ (−2, ∞)
A polynomial plus a fraction
Move every term to the left side.
x + 4x − 5 < 0
The LCD is x.
Write one fraction.
x² − 5x + 4x < 0
Factor the numerator.
(x − 4)(x − 1)x < 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 1, x = 4
Divide-by-zero values are never solutions.
x ≠ 0
Try x = −1.
The value is −10, which is negative.
Try x = 12.
The value is 72, which is positive.
Try x = 52.
The value is −910, which is negative.
Try x = 5.
The value is 45, which is positive.
(−∞, 0) ∪ (1, 4)
Answer: (−∞, 0) ∪ (1, 4)
Move every term to the left side.
x + 6x − 5 ≤ 0
The LCD is x.
Write one fraction.
x² − 5x + 6x ≤ 0
Factor the numerator.
(x − 3)(x − 2)x ≤ 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 2, x = 3
Divide-by-zero values are never solutions.
x ≠ 0
These numerator zeros are solutions because the symbol includes equals.
x = 2, x = 3
Try x = −1.
The value is −12, which is negative.
Try x = 1.
The value is 2, which is positive.
Try x = 52.
The value is −110, which is negative.
Try x = 4.
The value is 12, which is positive.
(−∞, 0) ∪ [2, 3]
Answer: (−∞, 0) ∪ [2, 3]
Move every term to the left side.
x + 8x − 6 > 0
The LCD is x.
Write one fraction.
x² − 6x + 8x > 0
Factor the numerator.
(x − 4)(x − 2)x > 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 2, x = 4
Divide-by-zero values are never solutions.
x ≠ 0
Try x = −1.
The value is −15, which is negative.
Try x = 1.
The value is 3, which is positive.
Try x = 3.
The value is −13, which is negative.
Try x = 5.
The value is 35, which is positive.
(0, 2) ∪ (4, ∞)
Answer: (0, 2) ∪ (4, ∞)
Move every term to the left side.
x + 3x − 4 ≥ 0
The LCD is x.
Write one fraction.
x² − 4x + 3x ≥ 0
Factor the numerator.
(x − 3)(x − 1)x ≥ 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 1, x = 3
Divide-by-zero values are never solutions.
x ≠ 0
These numerator zeros are solutions because the symbol includes equals.
x = 1, x = 3
Try x = −1.
The value is −8, which is negative.
Try x = 12.
The value is 52, which is positive.
Try x = 2.
The value is −12, which is negative.
Try x = 4.
The value is 34, which is positive.
(0, 1] ∪ [3, ∞)
Answer: (0, 1] ∪ [3, ∞)
Move every term to the left side.
x + 2x − 3 < 0
The LCD is x.
Write one fraction.
x² − 3x + 2x < 0
Factor the numerator.
(x − 2)(x − 1)x < 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 1, x = 2
Divide-by-zero values are never solutions.
x ≠ 0
Try x = −1.
The value is −6, which is negative.
Try x = 12.
The value is 32, which is positive.
Try x = 32.
The value is −16, which is negative.
Try x = 3.
The value is 23, which is positive.
(−∞, 0) ∪ (1, 2)
Answer: (−∞, 0) ∪ (1, 2)
Move every term to the left side.
x + 12x − 7 ≤ 0
The LCD is x.
Write one fraction.
x² − 7x + 12x ≤ 0
Factor the numerator.
(x − 4)(x − 3)x ≤ 0
Critical numbers come from the numerator and from the denominator.
x = 0, x = 3, x = 4
Divide-by-zero values are never solutions.
x ≠ 0
These numerator zeros are solutions because the symbol includes equals.
x = 3, x = 4
Try x = −1.
The value is −20, which is negative.
Try x = 32.
The value is 52, which is positive.
Try x = 72.
The value is −114, which is negative.
Try x = 5.
The value is 25, which is positive.
(−∞, 0) ∪ [3, 4]
Answer: (−∞, 0) ∪ [3, 4]
Two rational expressions
Move every term to the left side.
4x + 1 − 1x − 5 < 0
The LCD is (x + 1)(x − 5).
Write one fraction.
3x − 21(x + 1)(x − 5) < 0
Factor the numerator.
3(x − 7)(x + 1)(x − 5) < 0
Critical numbers come from the numerator and from the denominator.
x = −1, x = 5, x = 7
Divide-by-zero values are never solutions.
x ≠ −1 and x ≠ 5
Try x = −2.
The value is −277, which is negative.
Try x = 2.
The value is 53, which is positive.
Try x = 6.
The value is −37, which is negative.
Try x = 8.
The value is 19, which is positive.
(−∞, −1) ∪ (5, 7)
Answer: (−∞, −1) ∪ (5, 7)
Move every term to the left side.
1x − 2 − 3x + 4 ≤ 0
The LCD is (x − 2)(x + 4).
Write one fraction.
10 − 2x(x − 2)(x + 4) ≤ 0
Factor the numerator.
−2(x − 5)(x − 2)(x + 4) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = −4, x = 2, x = 5
Divide-by-zero values are never solutions.
x ≠ −4 and x ≠ 2
These numerator zeros are solutions because the symbol includes equals.
x = 5
Try x = −5.
The value is 207, which is positive.
Try x = −1.
The value is −43, which is negative.
Try x = 72.
The value is 415, which is positive.
Try x = 6.
The value is −120, which is negative.
(−4, 2) ∪ [5, ∞)
Answer: (−4, 2) ∪ [5, ∞)
Move every term to the left side.
3x + 2 − 1x − 6 ≤ 0
The LCD is (x + 2)(x − 6).
Write one fraction.
2x − 20(x + 2)(x − 6) ≤ 0
Factor the numerator.
2(x − 10)(x + 2)(x − 6) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = −2, x = 6, x = 10
Divide-by-zero values are never solutions.
x ≠ −2 and x ≠ 6
These numerator zeros are solutions because the symbol includes equals.
x = 10
Try x = −3.
The value is −269, which is negative.
Try x = 2.
The value is 1, which is positive.
Try x = 8.
The value is −15, which is negative.
Try x = 11.
The value is 265, which is positive.
(−∞, −2) ∪ (6, 10]
Answer: (−∞, −2) ∪ (6, 10]
Move every term to the left side.
6x + 3 − 2x − 1 < 0
The LCD is (x + 3)(x − 1).
Write one fraction.
4x − 12(x + 3)(x − 1) < 0
Factor the numerator.
4(x − 3)(x + 3)(x − 1) < 0
Critical numbers come from the numerator and from the denominator.
x = −3, x = 1, x = 3
Divide-by-zero values are never solutions.
x ≠ −3 and x ≠ 1
Try x = −4.
The value is −285, which is negative.
Try x = −1.
The value is 4, which is positive.
Try x = 2.
The value is −45, which is negative.
Try x = 4.
The value is 421, which is positive.
(−∞, −3) ∪ (1, 3)
Answer: (−∞, −3) ∪ (1, 3)
Move every term to the left side.
x + 4x − 2 − 3 ≤ 0
The LCD is (x − 2).
Write one fraction.
10 − 2x(x − 2) ≤ 0
Factor the numerator.
−2(x − 5)(x − 2) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = 2, x = 5
Divide-by-zero values are never solutions.
x ≠ 2
These numerator zeros are solutions because the symbol includes equals.
x = 5
Try x = 1.
The value is −8, which is negative.
Try x = 72.
The value is 2, which is positive.
Try x = 6.
The value is −12, which is negative.
(−∞, 2) ∪ [5, ∞)
Answer: (−∞, 2) ∪ [5, ∞)
Move every term to the left side.
2x − 2x + 3 − 1 ≤ 0
The LCD is (x + 3).
Write one fraction.
x − 5(x + 3) ≤ 0
Factor the numerator.
(x − 5)(x + 3) ≤ 0
Critical numbers come from the numerator and from the denominator.
x = −3, x = 5
Divide-by-zero values are never solutions.
x ≠ −3
These numerator zeros are solutions because the symbol includes equals.
x = 5
Try x = −4.
The value is 9, which is positive.
Try x = 1.
The value is −1, which is negative.
Try x = 6.
The value is 19, which is positive.
(−3, 5]
Answer: (−3, 5]
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