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roots question - difficult

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Source: — Problem Solving |

by sg1978 » Wed Jan 13, 2010 5:55 pm
IMO E
Multiply both the denominator and numerator by (root n+1) + (root n). Then the denominator
has the form of (a+b)(a-b)=a^2-b^2 and in this case evaluates to 1. So the numerator is
now (root n+1) + (root n). So answer is E.
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by ace_gre » Wed Jan 13, 2010 5:57 pm
Hi, My approach is to use the identity a^2 - b^2 = (a+b) (a-b)
Multiply both numerator and denominator with (root n+1) + (root n),
expression ==> (root n+1) + (root n) / (n+1-n). Denominator =1,
Hence IMO E
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by Testluv » Wed Jan 13, 2010 7:10 pm
In general on the GMAT, whenever we have a radical in the denominator, we have to "rationalize" the expression. It is considered bad form to have a radical in the denominator of an expression.

So, let's say you're working on a question, and you correctly solve it to x = 3/root 5. You won't see this expression among the answer choices. You have to rationalize it by multiplying it by 1. And the kind of "1) we multiply it by is root 5/root 5. So:

3/root 5 * (root 5/root 5) = 3root5/5...that's the answer that will show up among the answer choices.

If the denominator is of the form (root 5) - x, you rationalize it by multiplying by root 5 + x/root 5 +x.
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