Difference between revisions of "2021 AMC 10B Problems/Problem 15"
(→Solution) |
m (→Solution) |
||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | We square <math>x+\frac{1}{x}=\sqrt5</math> to get <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract 2 on both sides for <math>x^2+\frac{1}{x^2}=3</math> and square again, and see that <math>x^4+2+\frac{1}{x^4}=9</math> so <math>x^4+\frac{1}{x^4}=7</math>. We can divide our original expression of <math>x^11-7x^7+x^3</math> by <math>x^7</math> to get that it is equal to <math>x^7(x^4-7+\frac{1}{x^4})</math>. Therefore because <math>x^4+\frac{1}{x^4}</math> is 7, it is equal to <math>x^7(0)=\boxed{(B) 0}</math>. | + | We square <math>x+\frac{1}{x}=\sqrt5</math> to get <math>x^2+2+\frac{1}{x^2}=5</math>. We subtract 2 on both sides for <math>x^2+\frac{1}{x^2}=3</math> and square again, and see that <math>x^4+2+\frac{1}{x^4}=9</math> so <math>x^4+\frac{1}{x^4}=7</math>. We can divide our original expression of <math>x^{11}-7x^7+x^3</math> by <math>x^7</math> to get that it is equal to <math>x^7(x^4-7+\frac{1}{x^4})</math>. Therefore because <math>x^4+\frac{1}{x^4}</math> is 7, it is equal to <math>x^7(0)=\boxed{(B) 0}</math>. |
Revision as of 14:32, 11 February 2021
Problem
The real number 
satisfies the equation 
. What is the value of 
Solution
We square 
to get 
. We subtract 2 on both sides for 
and square again, and see that 
so 
. We can divide our original expression of 
by 
to get that it is equal to 
. Therefore because 
is 7, it is equal to 
.
