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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..
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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)

Karishma--

I understand the algebraic method of solving the problem, but I don't understand why plugging numbers isn't accurate on these kinds of problems. Could you explain? Thanks!
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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)

Karishma--

I understand the algebraic method of solving the problem, but I don't understand why plugging numbers isn't accurate on these kinds of problems. Could you explain? Thanks!

As explained by Bunuel above, plugging-in numbers isn't necessarily a problem. Sometimes, it can be problematic because multiple options may match. Then you would need to try another set of numbers on the options that matched. Usually, you will get one correct answer within two rounds.
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Hi Bunuel and Karisma,
does that mean square root of 4 is always 2 and not -2?
I thought it had two options just like x to the power 2 = 4 gives, x = 2 and x = -2.
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GMatAspirerCA
Hi Bunuel and Karisma,
does that mean square root of 4 is always 2 and not -2?
I thought it had two options just like x to the power 2 = 4 gives, x = 2 and x = -2.

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, \(\sqrt{4}=2\), NOT +2 or -2.

In contrast, the equation \(x^2=4\) has TWO solutions, +2 and -2.

Hope this helps.
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What's the difference between A and C?

Both will give us the same answer right?

A = - (of a positive xy)
B = - (of a positive xy)
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ameyaprabhu
What's the difference between A and C?

Both will give us the same answer right?

A = - (of a positive xy)
B = - (of a positive xy)

You are given that x < 0 (say x is -2) and y > 0 (say y is 5).
So xy will be -2*5 = -10.

Hence xy will actually be a negative number.

Option (A) = - xy = - (-10) = 10 (a positive number)

Option (C) = - |xy| = - |-10| = -10 (a negative number)
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VeritasPrepKarishma

am i correct in saying that \(\sqrt{x^{2}}\) is NOT the same as writing \(\sqrt{(x^{2})}\), so is this the reason here that \(\sqrt{x^{2}}\) = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line
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VeritasPrepKarishma

am i correct in saying that \(\sqrt{x^{2}}\) is NOT the same as writing \(\sqrt{(x^{2})}\), so is this the reason here that \(\sqrt{x^{2}}\) = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line

\(\sqrt{x^{2}}\) is the same as \(\sqrt{(x^{2})}\).

\(\sqrt{x^{2}}=|x|\). We are given that x is negative. When x < 0, we know that |x| = -x, so \(\sqrt{x^{2}}=|x|=-x\). Notice that since x is negative, -x = -negative = positive, thus both the square root and the absolute value return positive result.
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LakerFan24
VeritasPrepKarishma

am i correct in saying that \(\sqrt{x^{2}}\) is NOT the same as writing \(\sqrt{(x^{2})}\), so is this the reason here that \(\sqrt{x^{2}}\) = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line

In addition to what Bunuel said above, let me also add a line on how I explain this in words: You are right. An absolute value can never be negative. So |x| will never be negative. But it can be equal to -x. When? When x itself is negative. Note that a variable x CAN stand for a negative value too. So if x itself is negative, -x becomes POSITIVE. And that is the case in which |x| is equal to -x (which is positive).
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KarishmaB
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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)

Hi KarishmaB Bunuel, can't be write (x^2y^2)^1/2 =|x*y|. Why do we have to put x and y in separate modulus?
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ankitapugalia
KarishmaB
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)

Hi KarishmaB Bunuel, can't be write (x^2y^2)^1/2 =|x*y|. Why do we have to put x and y in separate modulus?

Both are the same: |xy| = |x|*|y|.
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This question is an example how even simple questions can go wrong if the basics are not clear
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