Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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| − | + | ==Problem== | |
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| + | The real number <math>x</math> satisfies the equation <math>x+\frac{1}{x} = \sqrt{5}</math>. What is the value of <math>x^{11}-7x^{7}+x^3?</math> | ||
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| + | <math>\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}</math> | ||
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| + | ==Solution== | ||
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| + | 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> sp <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 14:11, 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 
sp 
. We can divide our original expression of 
by 
to get that it is equal to 
. Therefore because 
is 7, it is equal to 
.
