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BukrsGmat
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For a negative integer x, what is the value of x*root(x^2)? Updated on: 18 Feb 2026, 21:14
Originally posted by BukrsGmat on 22 Jan 2012, 23:17.
Last edited by Bunuel on 18 Feb 2026, 21:14, edited 1 time in total.
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For a negative integer x, what is the value of x * √(x^2) ?

A) x^2
B) -x^2
C) x
D) -x
E) -1


Show Spoiler
In one Site I read the below fact. How true is it. Does Gmat provide this special condition.

√x : √ sign (the radical sign) is called the "Principal root" and always results in a non-negative number.
(x)^1/2: The x to the power of 1/2 is the result of the equation, y^2 = x. In this case the result could be "positive or negative or zero".

So here is the conclusion
√x : Always non-negative (note I am not saying its always "positive")
(x)^1/2: Can be positive or negative or zero
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For a negative integer x, what is the value of x*root(x^2)? 23 Jan 2012, 04:23
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sujit2k7
In one Site I read the below fact. How true is it. Does Gmat provide this special condition.

√x : √ sign (the radical sign) is called the "Principal root" and always results in a non-negative number.
(x)^1/2: The x to the power of 1/2 is the result of the equation, y^2 = x. In this case the result could be "positive or negative or zero".

So here is the conclusion
√x : Always non-negative (note I am not saying its always "positive")
(x)^1/2: Can be positive or negative or zero

Q. For a negative integer x, what is the value of x * √(x^2) ?
A) x^2
B) -x^2
C) x
D) -x
E) -1
Show more

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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For a negative integer x, what is the value of x*root(x^2)? 23 Jan 2012, 07:43
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sujit2k7
In one Site I read the below fact. How true is it. Does Gmat provide this special condition.

√x : √ sign (the radical sign) is called the "Principal root" and always results in a non-negative number.
(x)^1/2: The x to the power of 1/2 is the result of the equation, y^2 = x. In this case the result could be "positive or negative or zero".

So here is the conclusion
√x : Always non-negative (note I am not saying its always "positive")
(x)^1/2: Can be positive or negative or zero

Q. For a negative integer x, what is the value of x * √(x^2) ?
A) x^2
B) -x^2
C) x
D) -x
E) -1

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 \(x=-2<0\) --> \([m](-2)*\sqrt{(-2)^2}=(-2)*2=-4=-(-2)^2\).

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.
Show more




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
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For a negative integer x, what is the value of x*root(x^2)? 23 Jan 2012, 09:01
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manalq8
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
Show more

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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For a negative integer x, what is the value of x*root(x^2)? 23 Jan 2012, 11:32
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oh I see. thanks alot Bunuel. You rock
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For a negative integer x, what is the value of x*root(x^2)? 31 Jan 2012, 23:31
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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?
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For a negative integer x, what is the value of x*root(x^2)? 01 Feb 2012, 02:25
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bsaikrishna
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?
Show more

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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For a negative integer x, what is the value of x*root(x^2)? 01 Feb 2012, 03:43
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can we say?

(sqrt(x) )^2 = sqrt((x^2) ) = |x| ( for even powers and even roots)
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For a negative integer x, what is the value of x*root(x^2)? 01 Feb 2012, 03:54
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bsaikrishna
can we say?

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

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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For a negative integer x, what is the value of x*root(x^2)? 12 Mar 2012, 00:31
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sujit2k7
In one Site I read the below fact. How true is it. Does Gmat provide this special condition.

√x : √ sign (the radical sign) is called the "Principal root" and always results in a non-negative number.
(x)^1/2: The x to the power of 1/2 is the result of the equation, y^2 = x. In this case the result could be "positive or negative or zero".

So here is the conclusion
√x : Always non-negative (note I am not saying its always "positive")
(x)^1/2: Can be positive or negative or zero

Q. For a negative integer x, what is the value of x * √(x^2) ?
A) x^2
B) -x^2
C) x
D) -x
E) -1

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 \(x=-2<0\) --> \([m](-2)*\sqrt{(-2)^2}=(-2)*2=-4=-(-2)^2\).

Answer: B.
Show more

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 \(|x|=-x\) --> \(-x*|x|= -x*(-x)= x^2\).
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For a negative integer x, what is the value of x*root(x^2)? 12 Mar 2012, 01:06
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rohitgoel15

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 \(|x|=-x\) --> \(-x*|x|= -x*(-x)= x^2\)[/color].
Show more

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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For a negative integer x, what is the value of x*root(x^2)? 12 Mar 2012, 01:35
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Got it .. thanks Bunuel!
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For a negative integer x, what is the value of x*root(x^2)? 27 Nov 2013, 09:46
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BukrsGmat
In one Site I read the below fact. How true is it. Does Gmat provide this special condition.

√x : √ sign (the radical sign) is called the "Principal root" and always results in a non-negative number.
(x)^1/2: The x to the power of 1/2 is the result of the equation, y^2 = x. In this case the result could be "positive or negative or zero".

So here is the conclusion
√x : Always non-negative (note I am not saying its always "positive")
(x)^1/2: Can be positive or negative or zero

For a negative integer x, what is the value of x * √(x^2) ?
A) x^2
B) -x^2
C) x
D) -x
E) -1
Show more

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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For a negative integer x, what is the value of x*root(x^2)? 18 Feb 2026, 21:05
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