Inequalities

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by GMATGuruNY » Mon Jun 24, 2013 4:23 am
abhirup1711 wrote:52. If x is positive, is x>3?
a) (x-1)^2 > 4
b) (x-2)^2 > 9
Statement 1: (x-1)² > 4
Case 1: x-1 > 2
x > 3.

Case 2: x-1 < -2
x < -1.
Not possible, since x must be positive.

Since only Case 1 is viable, x > 3.
SUFFICIENT.

Statement 2: (x-2)² > 9
Case 3: x-2 > 3
x > 5.

Case 4: x-2 < -3
x < -1.
Not possible, since x must be positive.

Since only Case 3 is viable, x > 5.
SUFFICIENT.

The correct answer is D.
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by vipulgoyal » Tue Jun 25, 2013 4:13 am
Mitch please explain
here in solution why inequality changed, i believe its not multiplied by - both sides
x-1 < -2
x < -1

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by GMATGuruNY » Tue Jun 25, 2013 5:55 am
vipulgoyal wrote:Mitch please explain
here in solution why inequality changed, i believe its not multiplied by - both sides
x-1 < -2
x < -1
(x-1)² > 4.

When we solve with regard to the NEGATIVE square root of 4, the following is implied:
(x-1)² > (-2)²

Here, when we take the square root of each side, we are in effect DIVIDING each side by a negative value, so the direction of the inequality must change:
(x-1)² > (-2)²

(x-1)(x-1) > (-2)(-2)

(x-1)(x-1) / (x-1) < (-2)(-2) / (-2) Each side is divided by a negative value, so > changes to <.

x-1 < -2

x < -1.

In general, when an inequality includes a square root, flip the inequality when solving for the negative root:
If (x-1)² > 4, then either x-1 > 2 or x-1 < -2.
If (x+2)² < 9, then either x+2 < 3 or x+2 > -3.
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by vipulgoyal » Tue Jun 25, 2013 10:05 pm
thank you Mitch as always very well explained