Difference between revisions of "2021 AMC 10B Problems/Problem 15"
m (→Solution) |
(→Solution) |
||
Line 5: | Line 5: | ||
<math>\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}</math> | <math>\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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*} | ||
+ | (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*}</cmath> |
Revision as of 21:15, 11 February 2021
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can divide our original expression of by to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , but notice that this is also a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),