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Difference between revisions of "2021 AMC 10B Problems/Problem 15"

(→‎Solution)
(→‎Solution 2)
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Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, but notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers),  
 
Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, but notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers),  
 
<cmath> \begin{align*}  
 
<cmath> \begin{align*}  
−
(x^11)-7x^7+x^3 &= (x^10+x^9)-7x^7+x^3 \\
+
(x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\
 
&=(2x^9+x^8)-7x^7+x^3 \\
 
&=(2x^9+x^8)-7x^7+x^3 \\
 
&=(3x^8+2x^7)-7x^7+x^3 \\
 
&=(3x^8+2x^7)-7x^7+x^3 \\
Line 23: Line 23:
 
&=\boxed{(\textbf{B}) 0}
 
&=\boxed{(\textbf{B}) 0}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 +
~Lcz

Revision as of 21:16, 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 1

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 $x^7(0)=\boxed{(B) 0}$.

Solution 2

Multiplying both sides by $x$ and using the quadratic formula, we get $\frac{\sqrt{5} \pm 1}{2}$. We can assume that it is $\frac{\sqrt{5}+1}{2}$, but notice that this is also a solution the equation $x^2-x-1=0$, i.e. we have $x^2=x+1$. Repeatedly using this on the given (you can also just note Fibonacci numbers), \begin{align*} (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\ &=(2x^9+x^8)-7x^7+x^3 \\ &=(3x^8+2x^7)-7x^7+x^3 \\ &=(3x^8-5x^7)+x^3 \\ &=(-2x^7+3x^6)+x^3 \\ &=(x^6-2x^5)+x^3 \\ &=(-x^5+x^4+x^3) &=-x^3(x^2-x-1) &=\boxed{(\textbf{B}) 0} \end{align*}

~Lcz

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