For a negative integer x, what is the value of x*root(x^2)?

SOME NOTES:

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\). Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

As for \(\sqrt{x}\) and \(x^{\frac{1}{2}}\): they are the same.

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

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).


BACK TO THE ORIGINAL QUESTION:
For a negative integer \(x\), what is the value of \(x*\sqrt{x^2}\)?
A. x^2
B. -x^2
C. x
D. -x
E. -1

Given: \(x<0\). Question: \(x*\sqrt{x^2}=?\)

According to the above notes: \(x*\sqrt{x^2}=x*|x|\), since \(x<0\) then \(|x|=-x\) --> \(x*|x|=x*(-x)=-x^2\).

Or just substitute some negative \(x\), let \(x=-2<0\) --> \((-2)*\sqrt{(-2)^2}=(-2)*2=-4=-(-2)^2\).

Answer: B.

Hope it's clear.

P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.
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Banuel, I have a question.

why can't we cancel out the square root with the power of 2?? let me show u!
x * √(x^2)

we can do X* (sqrt x)^2, then we will be able to cancel the sqrt with power and we end up with x^2 which is ans A?

your explain is greatly appreciated. thanks

We have \(\sqrt{x^2}\) and not \((\sqrt{x})^2\). In our case, as given that \(x<0\), then \(\sqrt{x}\) would be undefined: even roots from negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers): \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{-25}=undefined\).

Notice that for \(x*\sqrt{x^2}\) we have: {negative number \(x\)}*{some positive value}={negative value}. Thus answer choices x^2 (A) and -x (D) are out right away as they are positive, -1 (E) is also out as given expression can not come down to a single numerical value. So we are left only with x^2 (B) and x (C), and as \(\sqrt{x^2}\) does not equal to 1 (\(x*\sqrt{x^2}\neq{x}\)), then x (C) is also out. So, we are left with the only possible correct answer x^2 (B).

Now, if it were: "For a POSITIVE integer \(x\), what is the value of \(x*(\sqrt{x})^2\)? (or \(x*\sqrt{x^2}\))" then the answer would be \(x^2\).
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oh I see. thanks alot Bunuel. You rock

Firstly, I would like to thank Bunuel for his excellent answers on this topic. Thanks a ton man!!

Okay, now I've got a small doubt here

Is sqrt(-2) undefined according to the GMAT?

If so, then what is the answer of ( sqrt(-2) )^2 ?

Is it -2 or is undefined or sth else?

I am no sure why we cannot cancel it and make the answer as -2, but at the same time a square cannot be negative?

An even root from a negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers):\(\sqrt[even]{negative}\) is undefined. So, \((\sqrt{-2})^2\) is undefined as the value under the square root is negative.

When you'll see \(\sqrt{expression}\) in realistic GMAT question then it would make clear that \(expression\geq{0}\).

Hope it's clear.
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can we say?

(sqrt(x) )^2 = sqrt((x^2) ) = |x| ( for even powers and even roots)

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

But again if you see:\((\sqrt{x})^2\) then there would me mentioned that \(x\geq{0}\) thus in this case \((\sqrt{x})^2=x\) right away.

Hope it's clear.
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Perfect explanation Bunuel .. I have a silly ques though.
Please pardon that by mistake I have marked some of your text in red

\(x*\sqrt{x^2}=x*|x|\), since \(x<0\) then \(|x|=-x\) --> \(x*|x|=x*(-x)=-x^2\).

Shouldn't this be as below considering x as -x as x < 0 :
\(x*\sqrt{x^2}= -x*|x|\), since \(x<0\) then \(|x|=-x\) --> \(-x*|x|= -x*(-x)= x^2\).

You have \(x\) there, why are you substituting it with \(-x\)? They are two different numbers.

Else you can plug some negative number to see the result, say \(x=-2\): \(x*\sqrt{x^2}=(-2)*\sqrt{(-2)^2}=(-2)*2=-4=-x^2\).

Hope it's clear.
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Got it .. thanks Bunuel!

Bunuel I still don't understand if we solve inside to out we have -2 whose square is 4 and then the roots for 4 are 2 and -2, since it has been given that X<0, hence we consider root as -2 which gives (-2) * (-2) ---> 4 that is X^2 then why is not the answer A...I have read that on GMAT we only consider positive roots, but then its not the same in DS questions...surely there is something I am missing ..Please can you point out that ...thx!!

Sorry!! read your explaination.. above its clear now ..Thanks a ton..so basically there is a thin line difference when we work with variables..

Bumping for review and further discussion.
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We are told x is a negative integer. Simply plug in -1 and solve:

x * √(x^2)
(-1)√((-1)^2)
(-1)(1)
-((-1)^2)

B

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