the sqrt() function must return a non-negative value.

(x-5)^2 = (5-x)^2. Either (x-5) or (5-x) is non-negative.

so in order for sqrt((x-5)^2) = (5 - x) to be true, 5-x must be non-negative and x must be less than 5.

statement 1:
-x|x|>0
so, -x must be greater than 0. therefore, x is negative. This satisfies our criteria for x<5. Sufficient.

statement 2:
5-x>0, again this is sufficient to prove the equation. Sufficient.

Choose D.

-BM-

Baldini wrote:Is sqrt((x-5)^2) = 5 - x?

1. -x|x|>0
2. 5-x>0

OA is D .

Can someone please explain how to get to this answer?
Thanks

The problem can be re-written as:
is |x -5| = 5 -x ?

This is based on the following rule:
sqrt(n^2) = |n|


stmt1: -x|x| > 0

This is true only if x -ve (ie. less than 0)
eg choose x= -3
-(-3) * |-3| = 3*3 = 9 which is > 0

Now plug in -3 into the Q stem.
|-3 -5| = 8
5 -(-3) = 8. So LHS = RHS

Sufficient.



Stmt2:
5 -x > 0
ie 5 > x => x < 5

let x = 4
|4-5| = 1
5 -4 = 1 OK

let x = 0
|0 -5| = 5 -0

Let x = -2
|-2 -5| = 5-(-2)
So sufficient.


So answer is D.

hope I dint mess up anywhere.