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If x is positive, then 1/(root(x+1) + root(x)) =

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If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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If x is positive, then \(\frac{1}{\sqrt{x+1}+\sqrt{x}}=\)

A) 1

B) x

C) \(\frac{1}{x}\)

D) \(\sqrt{x+1}-\sqrt{x}\)

E) \(\sqrt{x+1}+\sqrt{x}\)
[Reveal] Spoiler: OA

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Re: If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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New post 21 Nov 2013, 02:11
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registerincog wrote:
If x is positive, then \(\frac{1}{\sqrt{x+1}+\sqrt{x}}\)

A) 1
B) x
C) \(\frac{1}{x}\)
D) \(\sqrt{x+1}-\sqrt{x}\)
E) \(\sqrt{x+1}+\sqrt{x}\)


Rationalize the fraction by multiplying it by \(\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}\) and apply \((a+b)(a-b)=a^2-b^2\) to the denominator:

\(\frac{1}{\sqrt{x+1}+\sqrt{x}}*\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=\sqrt{x+1}-\sqrt{x}\).

Answer: D.
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Re: If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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New post 18 Dec 2013, 05:42
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Am I being stupid, or is the question missing something?

I.e. should there not be a question mark or something at the end, or at least an equal sign.

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Re: If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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New post 18 Dec 2013, 06:53

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Re: If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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New post 11 Sep 2014, 03:01
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registerincog wrote:
If x is positive, then \(\frac{1}{\sqrt{x+1}+\sqrt{x}}=\)

A) 1
B) x
C) \(\frac{1}{x}\)
D) \(\sqrt{x+1}-\sqrt{x}\)
E) \(\sqrt{x+1}+\sqrt{x}\)


For such type of questions, change the middle sign & reciprocal

For example, \(\frac{1}{\sqrt{6} + \sqrt{5}} = \sqrt{6} - \sqrt{5}\)

Answer = D
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Re: If x is positive, then 1/(root(x+1) + root(x)) = [#permalink]

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Re: If x is positive, then 1/(root(x+1) + root(x)) =   [#permalink] 05 Oct 2017, 07:21
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