PareshGmat
VeritasPrepKarishma
nehamodak
What is \sqrt{x^2*y^2} if x < 0 and y > 0?
(A) –xy
(B) xy
(C) –|xy|
(D) |y|x
(E) No solution
Please explain how do we solve this
Note that \(\sqrt{x^2} = |x|\). It is not x, it is |x|. When we talk about square root, it implies the principal square root i.e. just the positive square root.
\(\sqrt{x^2*y^2} = |x|*|y|\)
Now, if x < 0, \(|x| = -x\)
If y > 0, \(|y| = y\)
Hence, \(|x|*|y| = -x*y\)
Answer (A)
It seems there is a risk if we plugin numbers eg: -2 & 2 which may not give an undisputed answer as computed by Karishma above

Experts, have your say..
If you plug x=-2 and y=2, then \(\sqrt{x^2*y^2}=4\):
(A) –xy = 4. Match.
(B) xy = -4. Discard.
(C) –|xy| = -4. Discard.
(D) |y|x = -4. Discard.
So, plunging in gives the same answer.
One could discard B, C and D right away. The square root function cannot give negative result, while all these options are negative, which means that they cannot be correct answers:
(A) –xy = -(negative)(positive) = positive. OK.
(B) xy = (negative)(positive) =
negative. Discard.
(C) –|xy| = -positive =
negative. Discard.
(D) |y|x = positive*negative =
negative. Discard.
Hope it's clear.