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Re: if x>0 then 1/[v(2x) + vx] [#permalink]
15 Dec 2010, 01:43

Expert's post

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}}.

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. _________________

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?

Re: if x>0 then 1/[v(2x) + vx] [#permalink]
17 Mar 2012, 07:04

1

This post received KUDOS

Expert's post

1

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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.

Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]
20 Mar 2012, 20:26

3

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}} _________________

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 _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink]
01 Sep 2014, 18:31

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