udaymathapati wrote:

If x < 0, then \sqrt{-x} •|x|) is

A. -x

B. -1

C. 1

D. x

E. \sqrt{x}

1. If x<0, then \(\sqrt{-x*|x|}\) equals:A. \(-x\)

B. \(-1\)

C. \(1\)

D. \(x\)

E. \(\sqrt{x}\)

Remember: \(\sqrt{x^2}=|x|\).

The point here is that

square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:

If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);

If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:

\(\sqrt{x^2}=x\), if \(x\geq{0}\);

\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=|x|\)

Back to the original question:\(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\)

Or just substitute the value let \(x=-5<0\) --> \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x\).

Answer: A.

Hope it's clear.

Sorry Bunuel but the approach :\(\sqrt{-x* |x|}\) if we put this one ^2 then we have simply \(-x * |x|\). The latter is positive, so we have a quantity straight negative \(- X\).