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# If x>0 then 1/[(root(2x)+root(x)]=?

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If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  14 Dec 2010, 18:08
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If $$x>0$$ then $$\frac{1}{\sqrt{2x}+\sqrt{x}}=?$$

A. $$\frac{1}{\sqrt{3x}}$$

B. $$\frac{1}{2\sqrt{3x}}$$

C. $$\frac{1}{x\sqrt{2}}$$

D. $$\frac{\sqrt{2}-1}{\sqrt{x}}$$

E. $$\frac{1+\sqrt{2}}{\sqrt{x}}$$

I have a doubt about the ans. Can someone explain this.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Mar 2012, 04:30, edited 1 time in total.
Edited the question and the answer choices.
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Re: if x>0 then 1/[v(2x) + vx] [#permalink]  14 Dec 2010, 18:52
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.

Hi!

unless you're missing exponents, then the answer should be (A).

v(2x) is the same as 2vx. So, if we're adding 2vx + 1vx, we get 3vx.

Of course, v(3x) is the same as 3vx, so (A) is correct.

Are you sure you've reproduced the question correctly?
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Re: if x>0 then 1/[v(2x) + vx] [#permalink]  14 Dec 2010, 18:55
the question is correct but i guess the oa is wrong. thanks
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Re: if x>0 then 1/[v(2x) + vx] [#permalink]  14 Dec 2010, 19:00
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ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.

Square roots in denominator should remind you of difference of squares.
$$\frac{1}{{\sqrt{2x}+\sqrt{x}}}$$

Multiply and divide by $${\sqrt{2x}-\sqrt{x}$$

You get $$\frac{1}{{\sqrt{2x}+\sqrt{x}}}$$ * $$({\sqrt{2x}-\sqrt{x}})/({\sqrt{2x}-\sqrt{x}})$$

You get $$({\sqrt{2x}-\sqrt{x})/x$$
Taking root x common from the numerator, we get $$({\sqrt{2}-1)/\sqrt{x}$$

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kaplan GMAT Instructor Joined: 21 Jun 2010 Posts: 148 Location: Toronto Followers: 43 Kudos [?]: 149 [0], given: 0 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 14 Dec 2010, 19:08 Oh.. if "v" is supposed to be "root of", then my response is completely wrong. Please use the math fonts for roots (or say "root"). Math Expert Joined: 02 Sep 2009 Posts: 30362 Followers: 5082 Kudos [?]: 57184 [1] , given: 8808 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 15 Dec 2010, 01:43 1 This post received KUDOS Expert's post 2 This post was BOOKMARKED ajit257 wrote: Q. if x>0 then 1/[v(2x) + vx] = ? a. 1/v(3x) b.1/[2v(2x)] c.1/(xv2) d. v2-1/vx e. 1+v2/vx I have a doubt about the ans. Can someone explain this. ajit257 please make sure the stem and the answer choices are not ambiguous. As OA is given as D then I guess v means square root, so the question should be: If $$x>0$$ then $$\frac{1}{\sqrt{2x}+\sqrt{x}}=?$$ Factor out $$\sqrt{x}$$ --> $$\frac{1}{\sqrt{x}(\sqrt{2}+1)}$$ --> multiply and divide by $$\sqrt{2}-1$$ --> $$\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}$$. Answer: D. _________________ Senior Manager Joined: 28 Aug 2010 Posts: 254 Followers: 4 Kudos [?]: 284 [0], given: 11 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 15 Dec 2010, 04:16 firstly apologies.....well it was hard to analyse the question in the first place the way it was written...I would not have got it wrong if it was a sq rt....the ans choice led me to ask the question...but thanks guys at-least i got the right question now. _________________ Verbal:new-to-the-verbal-forum-please-read-this-first-77546.html Math: new-to-the-math-forum-please-read-this-first-77764.html Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html ------------------------------------------------------------------------------------------------- Ajit Intern Joined: 16 Mar 2012 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 16 Mar 2012, 04:22 VeritasPrepKarishma, thank you Manager Joined: 14 Dec 2011 Posts: 65 GMAT 1: 630 Q48 V29 GMAT 2: 690 Q48 V37 Followers: 1 Kudos [?]: 22 [0], given: 24 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 17 Mar 2012, 06:20 Bunuel wrote: ajit257 wrote: Q. if x>0 then 1/[v(2x) + vx] = ? Factor out $$\sqrt{x}$$ --> $$\frac{1}{\sqrt{x}(\sqrt{2}+1)}$$ --> multiply and divide by $$\sqrt{2}-1$$ --> $$\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}$$. Answer: D. Could you please explain why you multiply and divide by $$\sqrt{2}-1$$ ...whats the concept here? Thanks! Math Expert Joined: 02 Sep 2009 Posts: 30362 Followers: 5082 Kudos [?]: 57184 [1] , given: 8808 Re: if x>0 then 1/[v(2x) + vx] [#permalink] 17 Mar 2012, 07:04 1 This post received KUDOS Expert's post 2 This post was BOOKMARKED Impenetrable wrote: Bunuel wrote: ajit257 wrote: Q. if x>0 then 1/[v(2x) + vx] = ? Factor out $$\sqrt{x}$$ --> $$\frac{1}{\sqrt{x}(\sqrt{2}+1)}$$ --> multiply and divide by $$\sqrt{2}-1$$ --> $$\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}$$. Answer: D. Could you please explain why you multiply and divide by $$\sqrt{2}-1$$ ...whats the concept here? Thanks! It's called rationalisation and is performed to eliminate irrational expression in the denominator. For this particular case we are doing rationalisation by applying the following rule: $$(a-b)(a+b)=a^2-b^2$$. $$\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}$$. Hope it' clear. _________________ Senior Manager Joined: 18 Sep 2009 Posts: 359 Followers: 3 Kudos [?]: 245 [0], given: 2 Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] 20 Mar 2012, 09:30 Karishma can you please explain last step . how did you get root x in the numerator Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1596 Kudos [?]: 8930 [5] , given: 195 Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] 20 Mar 2012, 20:26 5 This post received KUDOS Expert's post TomB wrote: Karishma can you please explain last step . how did you get root x in the numerator $$\frac{1}{\sqrt{2x} + \sqrt{x}}*\frac{\sqrt{2x} - \sqrt{x}}{\sqrt{2x} - \sqrt{x}} = \frac{\sqrt{2x} - \sqrt{x}}{x} = \frac{\sqrt{x}(\sqrt{2} - 1)}{\sqrt{x}*\sqrt{x}}$$ (Take $$\sqrt{x}$$ common from the two terms of the numerator. Split the x in the denominator into $$\sqrt{x}*\sqrt{x}$$ Now cancel the $$\sqrt{x}$$ from the numerator and denominator. You get $$\frac{(\sqrt{2} - 1)}{\sqrt{x}}$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  23 Mar 2012, 03:54
thanks for the explanations. they are very helpful.
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  26 Jul 2012, 03:08
ajit257 wrote:
If $$x>0$$ then $$\frac{1}{\sqrt{2x}+\sqrt{x}}=?$$

A. $$\frac{1}{\sqrt{3x}}$$

B. $$\frac{1}{2\sqrt{3x}}$$

C. $$\frac{1}{x\sqrt{2}}$$

D. $$\frac{\sqrt{2}-1}{\sqrt{x}}$$

E. $$\frac{1+\sqrt{2}}{\sqrt{x}}$$

I have a doubt about the ans. Can someone explain this.

Just by rationalising the denominator we get D
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  01 Sep 2014, 18:31
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  01 Sep 2014, 20:21
rajathpanta wrote:
ajit257 wrote:
If $$x>0$$ then $$\frac{1}{\sqrt{2x}+\sqrt{x}}=?$$

A. $$\frac{1}{\sqrt{3x}}$$

B. $$\frac{1}{2\sqrt{3x}}$$

C. $$\frac{1}{x\sqrt{2}}$$

D. $$\frac{\sqrt{2}-1}{\sqrt{x}}$$

E. $$\frac{1+\sqrt{2}}{\sqrt{x}}$$

I have a doubt about the ans. Can someone explain this.

Just by rationalising the denominator we get D

Just adding a small note as all guys above have explained.......

Rationalising would change the middle sign in the problem for the numerator

Only option D fits in...
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  10 Mar 2015, 04:31
VeritasPrepKarishma wrote:
ajit257 wrote:

Taking root x common from the numerator, we get $$({\sqrt{2}-1)/\sqrt{x}$$

I don't understand this step, could you please explain?
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  10 Mar 2015, 04:38
Expert's post
erikvm wrote:
VeritasPrepKarishma wrote:
ajit257 wrote:

Taking root x common from the numerator, we get $$({\sqrt{2}-1)/\sqrt{x}$$

I don't understand this step, could you please explain?

$$\frac{\sqrt{2x}-\sqrt{x}}{x}=\frac{\sqrt{2}\sqrt{x}-\sqrt{x}}{x}=\frac{\sqrt{x}(\sqrt{2}-1)}{x}=\frac{(\sqrt{2}-1)}{\sqrt{x}}$$
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If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  10 Mar 2015, 10:48
I don't understand this step, could you please explain?[/quote]

$$\frac{\sqrt{2x}-\sqrt{x}}{x}=\frac{\sqrt{2}\sqrt{x}-\sqrt{x}}{x}=\frac{\sqrt{x}(\sqrt{2}-1)}{x}=\frac{(\sqrt{2}-1)}{\sqrt{x}}$$[/quote]

Hi, I don't quite get this last step. We are here:

multiply and divide by \sqrt{2}-1 --> \frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}.

Now how do we get to here: \frac{\sqrt{2}-1}{\sqrt{x}}.

?
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]  10 Mar 2015, 10:51
Bunuel wrote:
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.

ajit257 please make sure the stem and the answer choices are not ambiguous.

As OA is given as D then I guess v means square root, so the question should be:

If $$x>0$$ then $$\frac{1}{\sqrt{2x}+\sqrt{x}}=?$$

Factor out $$\sqrt{x}$$ --> $$\frac{1}{\sqrt{x}(\sqrt{2}+1)}$$ --> multiply and divide by $$\sqrt{2}-1$$ --> $$\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}$$.

What is the last step here? I do not quite understand it
Re: If x>0 then 1/[(root(2x)+root(x)]=?   [#permalink] 10 Mar 2015, 10:51

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