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

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Intern
Joined: 16 Nov 2013
Posts: 28
Location: United States
Concentration: Entrepreneurship, General Management
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If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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21 Nov 2013, 02:03
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Difficulty:

5% (low)

Question Stats:

80% (00:41) correct 20% (01:38) wrong based on 158 sessions

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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) $$\sqrt{x+1}+\sqrt{x}$$
Math Expert
Joined: 02 Sep 2009
Posts: 47948
Re: If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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21 Nov 2013, 02:11
2
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:

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

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

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18 Dec 2013, 05:42
1
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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Joined: 02 Sep 2009
Posts: 47948
Re: If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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18 Dec 2013, 06:53
wuhanmatt wrote:
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.

You are right there should be an equal sign at the end. Edited. Thank you.
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Re: If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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11 Sep 2014, 03:01
3
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, $$\frac{1}{\sqrt{6} + \sqrt{5}} = \sqrt{6} - \sqrt{5}$$

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Joined: 04 Jun 2018
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Re: If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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27 Jul 2018, 12:25
PareshGmat wrote:
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, $$\frac{1}{\sqrt{6} + \sqrt{5}} = \sqrt{6} - \sqrt{5}$$

I just encountered this question as part of my daily questions from the log and only got it right by guessing.

Could you elaborate how you get root 5&6?

Thanks and best regards,
Chris
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Joined: 02 Sep 2009
Posts: 47948
Re: If x is positive, then 1/(root(x+1) + root(x)) =  [#permalink]

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28 Jul 2018, 04:34
1
Arro44 wrote:
PareshGmat wrote:
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, $$\frac{1}{\sqrt{6} + \sqrt{5}} = \sqrt{6} - \sqrt{5}$$

I just encountered this question as part of my daily questions from the log and only got it right by guessing.

Could you elaborate how you get root 5&6?

Thanks and best regards,
Chris

This is explained HERE.

This algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator.

Questions involving rationalization to practice:
https://gmatclub.com/forum/in-the-given ... 59791.html
http://gmatclub.com/forum/if-x-0-then-106291.html
http://gmatclub.com/forum/if-n-is-posit ... 31236.html
http://gmatclub.com/forum/consider-a-qu ... 31083.html
http://gmatclub.com/forum/in-the-diagra ... 39282.html
http://gmatclub.com/forum/in-the-diagra ... 29962.html
http://gmatclub.com/forum/the-perimeter ... 27049.html
http://gmatclub.com/forum/which-of-the- ... 98531.html
http://gmatclub.com/forum/if-x-is-posit ... 63491.html
http://gmatclub.com/forum/1-2-sqrt3-64378.html
http://gmatclub.com/forum/if-a-square-m ... 99359.html

Hope it helps.
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Re: If x is positive, then 1/(root(x+1) + root(x)) = &nbs [#permalink] 28 Jul 2018, 04:34
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