Difference between revisions of "1976 AHSME Problems/Problem 27"

(Created page with "== Problem 27 == If <math>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}</math>, then <math>N</math> equals <math>\textbf{(A) }1\qquad \t...")
 
(Problem 27)
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\textbf{(D) }\sqrt{\frac{5}{2}}\qquad
 
\textbf{(D) }\sqrt{\frac{5}{2}}\qquad
 
\textbf{(E) }\text{none of these}  </math>
 
\textbf{(E) }\text{none of these}  </math>
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==Solution==
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We will split this problem into two parts: The fraction on the left and the square root on the right.
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Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes <math>\frac{\sqrt{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}}{\sqrt{\sqrt{5}+1}}</math>

Revision as of 17:56, 20 March 2020

Problem 27

If $N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$, then $N$ equals

$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$

Solution

We will split this problem into two parts: The fraction on the left and the square root on the right.

Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes $\frac{\sqrt{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}}{\sqrt{\sqrt{5}+1}}$

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