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x^2 – x – 6 < 0?

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by sanju09 » Sat Oct 22, 2011 3:53 am
What is the solution set of the inequality x^2 – x – 6 < 0?
(A) x > –2
(B) –2 < x < 3
(C) x > 3 and x < –2
(D) x > 3 or x < –2
(E) x < 3
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by Anurag@Gurome » Sat Oct 22, 2011 4:08 am
sanju09 wrote:What is the solution set of the inequality x^2 – x – 6 < 0?
--> x² - x - 6 < 0
--> (x - 3)(x + 2) < 0
--> -2 < x < 3

The correct answer is B.

If anyone needs more clarification:
  • (x - 3)(x + 2) < 0 implies that (x - 3) and (x + 2) will have opposite signs. Hence, any one of the following is possible,
    • 1. (x - 3) < 0 and (x + 2) > 0
      -> x < 3 and x > -2
      -> -2 < x < 3
    • 2. (x - 3) > 0 and (x + 2) < 0
      -> x > 3 and x < -2
      -> No such x is possible
Hence, overall possible range for x is -2 < x < 3
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by neelgandham » Sat Oct 22, 2011 8:31 am
Correct answer = [spoiler]-2 < x < 3 Option B[/spoiler].

How ?

We can simplify the equation to (x+2)(x-3). From the equation we know that it changes it's sign at x = -2 and x = 3, So we have three limits to check for

a) When x < -2
b) When x > -2 and x < 3 =>
c) When x > 3

If x < -2 then the equation(x+2)(x-3) is > than 0
If -2<x<3 then the equation(x+2)(x-3) is < than 0
If x > 3 then the equation(x+2)(x-3) is > than 0
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by rijul007 » Sat Oct 22, 2011 10:38 am
x^2 - x - 6 < 0
x^2 - 3x + 2x - 6 < 0
x(x - 3) + 2(x - 3) < 0
(x + 2)(x - 3) < 0

Image

-2 < x < 3

Option B
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