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Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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==Solution==
 
==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>.
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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> 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 14:11, 11 February 2021

Problem

The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$. What is the value of $x^{11}-7x^{7}+x^3?$

$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$

Solution

We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$. We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$. We can divide our original expression of $x^11-7x^7+x^3$ by $x^7$ to get that it is equal to $x^7(x^4-7+\frac{1}{x^4})$. Therefore because $x^4+\frac{1}{x^4}$ is 7, it is equal to $x^7(0)=\boxed{(B) 0}$.

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