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exponents and conjugate

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by Gurpinder » Tue Aug 24, 2010 9:49 am
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Hey guys,

I know for something like this one, you cannot have roots in the denominator, so the easiest way to get rid of roots in the denominator is to multiply it by itself and multiply the top by the same.

However, when you multiply do you keep the same signs or opposite?

Thanks,
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by 4GMAT_Mumbai » Tue Aug 24, 2010 10:36 am
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.
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by Maciek » Tue Aug 24, 2010 10:36 am
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!
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by Gurpinder » Tue Aug 24, 2010 10:41 am
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?
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by 4GMAT_Mumbai » Wed Aug 25, 2010 12:11 pm
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.
Naveenan Ramachandran
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