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

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

Revision as of 21:16, 11 February 2021

Problem

Solution 1

Solution 2

@@ Line 19: / Line 19: @@
 &=(-2x^7+3x^6)+x^3 \\
 &=(x^6-2x^5)+x^3 \\
-&=(-x^5+x^4+x^3)
+&=(-x^5+x^4+x^3) \\
-&=-x^3(x^2-x-1)
+&=-x^3(x^2-x-1) = \boxed{(\textbf{B}) 0}
-&=\boxed{(\textbf{B}) 0}
 \end{align*}</cmath>
 ~Lcz