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DS - Inequality

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DS - Inequality

by karthikpandian19 » Fri Jul 06, 2012 4:20 pm
Is x > -3?

x2 - 9 > x + 3
|x3| < 64
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by eagleeye » Fri Jul 06, 2012 5:04 pm
karthikpandian19 wrote:Is x > -3?

x2 - 9 > x + 3
|x3| < 64
We need to find whether x>-3

With that in mind, let's look at statement 1.
x^2-9 > x+3
x^2-3^2 > x+3
(x+3)*(x-3) > (x+3)
(x+3)*(x-3) - (x+3) > 0
(x+3)*(x-3-1)>0
(x+3)*(x-4)>0
From these equations, we get that x < -3 or x > 4. Insufficient.

Statement 2 reads:
|x^3|<64
|x^3|<4^3
|x|<4
-4 < x < 4. x lies between -4 and 4. We don't know whether x > -3 or not. Insufficient.

Together the statements are (x <-3 or x>4 ) AND -4 < x < 4
The only common region is -4 < x < -3, where the two statements are satisfied simultaneously. Hence x < -3, so we know for sure that x is not greater than -3. Sufficient.

Hence C is the answer.

Let me know if this helps :)
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by Bill@VeritasPrep » Fri Jul 06, 2012 5:47 pm
Is x > -3?

x2 - 9 > x + 3
|x3| < 64

Statement 1

x2 - 9 > x + 3 (move everything to the left side)

x2 - x - 9 - 3 > 0 (simplify)

x2 - x -12 > 0 (factor by inspection)

(x+3)(x-4) > 0

Either x < -3 or x > 4; insufficient

Statement 2

|x3| < 64

|x3| < 4^3

|x| < 4

x is between -4 and 4; insufficient.

Combined

The only possible values of x that satisfy both conditions are between -4 and -3. Sufficient for us to answer definitely yes.
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