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![If [(x + 1)(x - 2)] / (x - 3) < 0, which of the following specifies](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If [(x + 1)(x - 2)] / (x - 3) < 0, which of the following specifies"){kind=icon}
20 Nov 2025, 22:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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66% (01:42) correct
34%
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If \(\frac{(x + 1)(x – 2) }{ x – 3}\) < 0, which of the following specifies all the possible values of \(x\)?
A. \(x\) < –1
B. 2 < \(x\) < 3
C. –1 < \(x\) < 3
D. \(x\) < –1 or 3 < \(x\)
E. \(x\) < –1 or 2 < \(x\) < 3
A. \(x\) < –1
B. 2 < \(x\) < 3
C. –1 < \(x\) < 3
D. \(x\) < –1 or 3 < \(x\)
E. \(x\) < –1 or 2 < \(x\) < 3
20 Nov 2025, 22:43
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Multiply (x-3)^2 on both sides:
(x+1)(x-2)(x-3) < 0
Using the wavy-curve approach:
x < -1 or 2 < x < 3
(E)
(x+1)(x-2)(x-3) < 0
Using the wavy-curve approach:
x < -1 or 2 < x < 3
(E)
21 Nov 2025, 01:23
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ExpertsGlobal5
To solve the inequality \(\frac{(x+1)(x-2)}{x-3} < 0\), we first identify the critical points. These are the values where the numerator is zero or the denominator is zero.
Critical Points:
1. \(x + 1 = 0 \implies x = -1\)
2. \(x - 2 = 0 \implies x = 2\)
3. \(x - 3 = 0 \implies x = 3\)
These points divide the number line into four intervals. We need to test the sign of the expression in each interval to see if it is negative (< 0).
Interval Analysis:
1. Interval \(x < -1\) (Test value: \(x = -2\))
\(\frac{(-2 + 1)(-2 - 2)}{-2 - 3} = \frac{(-)(-)}{(-)} = \frac{\text{positive}}{\text{negative}} = \text{Negative}\)
This interval works.
2. Interval \(-1 < x < 2\) (Test value: \(x = 0\))
\(\frac{(0 + 1)(0 - 2)}{0 - 3} = \frac{(+)(-)}{(-)} = \frac{\text{negative}}{\text{negative}} = \text{Positive}\)
This interval does not work.
3. Interval \(2 < x < 3\) (Test value: \(x = 2.5\))
\(\frac{(2.5 + 1)(2.5 - 2)}{2.5 - 3} = \frac{(+)(+)}{(-)} = \frac{\text{positive}}{\text{negative}} = \text{Negative}\)
This interval works.
4. Interval \(x > 3\) (Test value: \(x = 4\))
\(\frac{(4 + 1)(4 - 2)}{4 - 3} = \frac{(+)(+)}{(+)} = \text{Positive}\)
This interval does not work.
Conclusion:
The inequality holds true for \(x < -1\) or \(2 < x < 3\).
Answer: E
