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If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ?

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hardnstrong
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If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   11 Jun 2010, 09:32
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If x ≠ 0, what is the value of \(\sqrt{\frac{x^u}{x^v}}\) ?

(1) u = v
(2) x is a perfect square
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A
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Bunuel
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If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   11 Jun 2010, 10:46
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If x ≠ 0, what is the value of \(\sqrt{\frac{x^u}{x^v}}\) ?

\(\sqrt{\frac{x^u}{x^v}}=?\)


(1) u=v --> \(\sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1\). Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   11 Jun 2010, 21:41
Bunuel wrote:
hardnstrong wrote:
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?
(1) u = v
(2) x is a perfect square.

OA will be posted later


\(\sqrt{\frac{x^u}{x^v}}=?\)

(1) \(u=v\) --> \(\sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1\). Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.


my question is sq. root of 1 can be +1 or -1 (like sq, root of 4 can be +2 or -2)
How can it be only 1?
please explain

OA is A
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   12 Jun 2010, 04:33
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hardnstrong wrote:
Bunuel wrote:
hardnstrong wrote:
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?
(1) u = v
(2) x is a perfect square.

OA will be posted later


\(\sqrt{\frac{x^u}{x^v}}=?\)

(1) \(u=v\) --> \(\sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1\). Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.


my question is sq. root of 1 can be +1 or -1 (like sq, root of 4 can be +2 or -2)
How can it be only 1?
please explain

OA is A


Square root function cannot give negative result.

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, +5 and -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 \(\sqrt[3]{-64} =-4\).

Hope it helps.
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   12 Jun 2010, 22:18
Didnt know that
Thanks for the precious info.
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   15 May 2017, 11:21
Bunuel wrote:
If x ≠ 0, what is the value of \(\sqrt{\frac{x^u}{x^v}}\) ?

(1) u = v
(2) x is a perfect square


stmt-1:
if u=v then all you have is what is the value of \sqrt{1}.
hence \(\sqrt{\frac{x^u}{x^v}}\) = 1
suffiecient

stmt-2:
say x = 4 but u can be 2 and v can be 4 and vice versa. this way we get two different values of \(\sqrt{\frac{x^u}{x^v}}\)

Answer should be A.
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink] New post   31 May 2017, 04:31
Statement 1 gives the answer straight forward as 1.

Statement 2 tell us that square root can be removed since x is a square but we don't know the value of u and v which can give multiple answers.

Option A is correct.
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Re: If x ≠ 0, what is the value of (x^u/x^v)^(1/2) ? [#permalink]
31 May 2017, 04:31
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Bunuel
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