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

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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 | 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}} = \frac{\sqrt{(\sqrt{5}+2)+(\sqrt{5}-2)+2( | |

− | \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} | + | \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{2\sqrt5+2\sqrt{(\sqrt5)^2-2^2}}}{\sqrt{\sqrt{5}+1}} = |

+ | |||

+ | \frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} |

## Revision as of 18:06, 20 March 2020

## Problem 27

If , then equals

## 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}} = \frac{\sqrt{(\sqrt{5}+2)+(\sqrt{5}-2)+2( \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{2\sqrt5+2\sqrt{(\sqrt5)^2-2^2}}}{\sqrt{\sqrt{5}+1}} =

\frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}}