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

Revision as of 15:02, 12 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 $\frac{\sqrt{5}+1}{2}$ , and 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

Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials)

https://youtu.be/hzcSPVGFbC8

~ pi_is_3.14

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 24: / Line 24: @@
 ~Lcz
+== Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials) ==
+https://youtu.be/hzcSPVGFbC8
+~ pi_is_3.14
 {{AMC10 box|year=2021|ab=B|num-b=14|num-a=16}}

Page

Toolbox

Search