maheshsrini wrote:

What is the value of |x|

(1) |x^2 + 16| – 5 = 27

(2) x^2 = 8x – 16

Love seeing the modes and the inequalities surrounding it , i often used to be scared until one day i realised what exactly is a inequality

Something that is not equal and a new world was born but the problem was when you mix absolute values in it and that is scary ,lol

Coming to the question, I would like to share a shortcut

Whenever you have a | x + a| = c , go ahead with x+a =+/- c and solve for each possibility of positive and negative

This way , you will save yourself lot of time

But then again revisiting basics , we know that square of a number can never be negative and lets see how that evolves out in this question prompt

(1) |x^2 + 16| – 5 = 27

can be written as |x^2 + 16|= 32

(a) Positive value

using our trick for positive value x^2 + 16 = 32

So , x^2 = 16

So x^2 - 16 = 0

Voila isn't the sweet boy looking familiar ...

Let me share another concept with you ...this is a equation with degree 2 , so it will have 2 factors

x^2 = 16 doesn't mean x=4 ...it mean there are two identical factors each of which is 4

Likewise whenever you have a equation having 3 degrees , always know that it shall have 3 factors

But anyways , if both value of x is 4 , isn't |x|=4 ....

easy to understand always know that |number| is distance of that number from zero on number line ...

So quite obviously |4|=4 ..

But wait , what about our second choice

(a) Negative value

There are two reasons it wont work

firstly , x^2 is positive and secondly 16 is positive as well

So when you add two positive numbers , GMAT will punish you if you show RHS to be negative

But say you do x^2 + 16 = -32

X^2=-48

Can square of a number be positive 1^2=1 , -1^2=1 ..

In fact square and mode are those sweet little devils which hide the sign and therefore when you have them , you actually dunno what the number is

But anyways , if you have square of a number , rest assured that it's cant be negative

And so my friend , we have to reject this solution

And so pick up the positive solution and therein lies our sufficiency for this part

So , hmmmm...lets see A is alive because statement 1 is sufficient , we can bid adieu to C,E( but they are still nice fellows ..maybe we need them later but not for this question , sowwie !)

(2) Statement 2

-x^2 = 8x – 16

Rearrange it a bit and it boils down to again (x-4)^2 = 0 again two roots and each of them 4

So , mode of 4 is 4 and sweet little boy D is our answer here