Hi,

If the denominator is of the format (a-b); multiply it with (a+b) and viceversa.

Only then, will the product be a^2 - b^2.

In this case, multiply the numerator and denominator by root(n+1) + root(n)

Hope this helps. Thanks.

Hi Gurpinder

my suggestion is:

try different approach and look at the answers first

E) √(n +1) + √n

(√(n +1) + √n) * (√(n +1) - √n)/(√(n +1) - √n) =
= (√(n +1)*√(n +1) - √(n +1) *√n + √n*√(n +1) - √n*√n)/(√(n +1) - √n) =
= (n + 1 - n)/(√(n +1) - √n) = 1/(√(n +1) - √n)

IMO E

hope it helps!

4GMAT_Mumbai wrote:Hi,

If the denominator is of the format (a-b); multiply it with (a+b) and viceversa.

Only then, will the product be a^2 - b^2.

In this case, multiply the numerator and denominator by root(n+1) + root(n)

Hope this helps. Thanks.

So basically my aim is to maintain the original signs.

so if the question is 1/sqrt[x-y] ----> i would multiply the denominator with sqrt[x+y] because that will keep the minus sign?

Right?

Gurpinder wrote:
So basically my aim is to maintain the original signs.

so if the question is 1/sqrt[x-y] ----> i would multiply the denominator with sqrt[x+y] because that will keep the minus sign?

Right?

Well, the function 'sqrt' is like a clasp on (x-y). So, if one wants to get rid of sqrt in the denominator, one has to multiply and divide by sqrt[x-y] (and not sqrt[x+y]).

The fraction becomes sqrt[x-y] / (x-y).

Hope this helps. Thanks.