Difference between revisions of "1977 AHSME Problems/Problem 7"
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| + | == Problem 7 == | ||
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| + | If <math>t = \frac{1}{1 - \sqrt[4]{2}}</math>, then <math>t</math> equals | ||
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| + | <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> | ||
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==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 ![$t = \frac{1}{1 - \sqrt[4]{2}}$](http://latex.artofproblemsolving.com/1/4/f/14ffcfccd5c911ca4a651a119adfc96cfe6c300b.png)
, then 
equals
![$\text{(A)}\ (1-\sqrt[4]{2})(2-\sqrt{2})\qquad \text{(B)}\ (1-\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(C)}\ (1+\sqrt[4]{2})(1-\sqrt{2}) \qquad \\ \text{(D)}\ (1+\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})$](http://latex.artofproblemsolving.com/6/b/7/6b7fa7a3892b258d4285b47b1daa7e78cbc48df0.png)
Solution
Solution by e_power_pi_times_i
![$t = \dfrac{1}{1-\sqrt[4]{2}} = (\dfrac{1}{1-\sqrt[4]{2}})(\dfrac{1+\sqrt[4]{2}}{1+\sqrt[4]{2}}) = (\dfrac{1+\sqrt[4]{2}}{1-\sqrt{2}})(\dfrac{1+\sqrt{2}}{1+\sqrt{2}}) = \dfrac{(1+\sqrt[4]{2})(1+\sqrt{2})}{-1} = \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})$](http://latex.artofproblemsolving.com/c/6/e/c6ede94c008ed200fa9a2ee4f8df9e437a7ff54b.png)
.
