Difference between revisions of "1977 AHSME Problems/Problem 7"
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+ | == Problem 7 == | ||
+ | |||
+ | If <math>t = \frac{1}{1 - \sqrt[4]{2}}</math>, then <math>t</math> equals | ||
+ | |||
+ | <math>\text{(A)}\ (1-\sqrt[4]{2})(2-\sqrt{2})\qquad | ||
+ | \text{(B)}\ (1-\sqrt[4]{2})(1+\sqrt{2})\qquad | ||
+ | \text{(C)}\ (1+\sqrt[4]{2})(1-\sqrt{2}) \qquad \\ | ||
+ | \text{(D)}\ (1+\sqrt[4]{2})(1+\sqrt{2})\qquad | ||
+ | \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2}) </math> | ||
+ | |||
+ | |||
+ | |||
==Solution== | ==Solution== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
<math>t = \dfrac{1}{1-\sqrt[4]{2}} = (\dfrac{1}{1-\sqrt[4]{2}})(\dfrac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}) = (\dfrac{1+\sqrt[4]{2}}{1-\sqrt{2}})(\dfrac{1+\sqrt{2}}{1+\sqrt{2}}) = \dfrac{(1+\sqrt[4]{2})(1+\sqrt{2})}{-1} = \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})</math>. | <math>t = \dfrac{1}{1-\sqrt[4]{2}} = (\dfrac{1}{1-\sqrt[4]{2}})(\dfrac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}) = (\dfrac{1+\sqrt[4]{2}}{1-\sqrt{2}})(\dfrac{1+\sqrt{2}}{1+\sqrt{2}}) = \dfrac{(1+\sqrt[4]{2})(1+\sqrt{2})}{-1} = \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})</math>. |
Latest revision as of 12:29, 21 November 2016
Problem 7
If , then
equals
Solution
Solution by e_power_pi_times_i
.