gmatter0913 wrote:

Hi Karishma,

I tried the problem x-1 < sqrt (7-x) as below:

As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)

x-1 can be -ve or +ve

When x-1<=0; x<=1 --------> (2)

When x-1>=0; x>=1 --------> (3)

As both sides are +ve, we can square both the sides

(x-1)^2 < 7-x

x^2 -x -6<0

(x-3)(x+2)<0

-2<x<3 ------------>(4)

The answer to this problem is (x<3). I am not sure how to arrive at that from hereon. Could you please help me?

You have done the process correctly. Now you need to understand what this implies.

You got x <= 7

Case 1: x-1< 0

When x-1< 0; x < 1

Note that when x-1 is negative, it will always be less than

\sqrt{(7-x)}So whenever x<1, the inequality will always hold.

Case 2: x-1 >= 0

If x-1 is non negative, we can square the inequality.

From this, you get -2<x<3.

The inequality holds in this range.

From the two cases, we see that the inequality holds for the range x < 3.

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Karishma

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