# 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...") |
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− | == | + | ==Solution== |

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

− | <math>\ | + | 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}} = \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}} = |

+ | \frac{\sqrt{2\sqrt5+2}}{\sqrt{\sqrt5+1}} = \frac{(\sqrt{2})(\sqrt{\sqrt5+1})}{\sqrt{\sqrt5+1}} = \sqrt2</math>. | ||

+ | |||

+ | Now for the right side. | ||

+ | |||

+ | <math>\sqrt{3-2\sqrt2} = \sqrt{(1-\sqrt2)^2} = 1-\sqrt2</math> | ||

+ | |||

+ | Putting it all together gives: | ||

+ | |||

+ | <math>(\sqrt2)+(1-\sqrt2)=\boxed{(A) 1}.</math> | ||

+ | |||

+ | ~Someonenumber011 |

## Latest revision as of 18:57, 20 March 2020

## 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

.

Now for the right side.

Putting it all together gives:

~Someonenumber011