Difference between revisions of "1976 AHSME Problems/Problem 27"
(→Solution) |
(→Solution) |
||
| Line 15: | Line 15: | ||
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 | ||
| − | + | <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}} = | \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}}</math> | |
| − | \frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} | ||
Revision as of 17:37, 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
