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Which of the following is equal to
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14 Sep 2017, 07:07
14 Sep 2017, 07:07
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Which of the following is equal to \(\frac{-2}{\sqrt{n-1} - \sqrt{n+1}}\) for all values of n > 1?
A. -1
B. 1
C. \(2(\sqrt{n-1} + \sqrt{n+1})\)
D. \(\sqrt{n-1} + \sqrt{n+1}\)
E. \(\frac{\sqrt{n-1}}{\sqrt{n+1}}\)
Re: Which of the following is equal to
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18 Sep 2017, 01:22
18 Sep 2017, 01:22
2
One way to simplify an expression containing square roots at the denominator is to proceed by rationalization. The dea is to multiply and divide the expression for the same number in order to eliminate the roots at the denominator. In our case,\(\frac{-2}{sqrt(n-1)-sqrt(n+1)}\) should be multiplied by \(\frac{sqrt(n-1)+sqrt(n+1)}{sqrt(n-1)+sqrt(n+1)}\) so that at the denominator we get a difference of squares.
Thus, \(\frac{-2}{sqrt(n-1)-sqrt(n+1)}*\frac{sqrt(n-1)+sqrt(n+1)}{sqrt(n-1)+sqrt(n+1)}=\frac{-2*[sqrt(n-1)+sqrt(n+1)]}{(n-1)-(n+1)}=\frac{-2*[sqrt(n-1)+sqrt(n+1)]}{-2}=sqrt(n-1)+sqrt(n+1)\).
Thus, the answer is D!
Thus, \(\frac{-2}{sqrt(n-1)-sqrt(n+1)}*\frac{sqrt(n-1)+sqrt(n+1)}{sqrt(n-1)+sqrt(n+1)}=\frac{-2*[sqrt(n-1)+sqrt(n+1)]}{(n-1)-(n+1)}=\frac{-2*[sqrt(n-1)+sqrt(n+1)]}{-2}=sqrt(n-1)+sqrt(n+1)\).
Thus, the answer is D!
Re: Which of the following is equal to
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04 Nov 2024, 16:03
04 Nov 2024, 16:03
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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