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If x 0, what is the value of sq. root (x^u/x^v) ?

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If x 0, what is the value of sq. root (x^u/x^v) ? [#permalink]

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11 Jun 2010, 09:32
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If x ≠ 0, what is the value of sq. root (x^u/x^v) ?

(1) u = v
(2) x is a perfect square.
[Reveal] Spoiler: OA

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Schools: IU, UT Dallas, Univ of Georgia, Univ of Arkansas, Miami University
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11 Jun 2010, 09:59
IMO E

A is INSUFFICIENT.
Equation becomes sqrt (x^(u-v)). If u-v = 0, then the value of the equation is +1 or -1.

B is INSUFFICIENT. We only know that x is a perfect square. Also, we know nothing about u & v.

A and B together dont provide us with any new information.

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11 Jun 2010, 10:46
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.

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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.

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?

OA is A
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Kudos [?]: 38 [0], given: 8

Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 129179 [0], given: 12194

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12 Jun 2010, 04:33
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.

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?

OA is A

Square root function can not 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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12 Jun 2010, 22:18
Didnt know that
Thanks for the precious info.
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Re: problem   [#permalink] 12 Jun 2010, 22:18
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