Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 April MIMC 10 Problems/Problem 12"

(Created page with "Given that <math>x^2-\frac{1}{x^2}=2</math>, what is <math>x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}</math>? <math>\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\text...")
 
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 3: Line 3:
 
<math>\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\textbf{(C)} ~3780 \qquad\textbf{(D)} ~840\sqrt{2} \qquad\textbf{(E)} ~1260\sqrt{2}</math>
 
<math>\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\textbf{(C)} ~3780 \qquad\textbf{(D)} ~840\sqrt{2} \qquad\textbf{(E)} ~1260\sqrt{2}</math>
 
==Solution==
 
==Solution==
−
To be Released on April 26th.
+
We know that <math>2=x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})</math>. Squaring both sides, we can get that <math>(x+\frac{1}{x})^2\cdot(x+\frac{1}{x})^2=4</math>. Therefore, <math>(x^2+2+\frac{1}{x^2})(x^2-2+\frac{1}{x^2})=4</math>. Setting <math>a=x^2+\frac{1}{x^2}</math>, we can get that <math>(a+2)(a-2)=4</math>, and therefore <math>a^2-4=4</math>. So <math>x^2+\frac{1}{x^2}=\sqrt{8}=2\sqrt{2}</math>. In addition, <math>x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=2\cdot2\sqrt{2}=4\sqrt{2}</math>. <math>\frac{2}{x^2}=2\sqrt{2}-2</math>, so <math>\frac{1}{x^2}=\sqrt{2}-1</math> and <math>\frac{1}{x^4}=3-2\sqrt{2}</math>. Therefore, <math>x^4+\frac{1}{x^4}=4\sqrt{2}+2(3-2\sqrt{2})=6</math>. Using a similar approach, we can calculate <math>x^8-\frac{1}{x^8}=4\sqrt{2}\cdot6=24\sqrt{2}</math> and <math>\frac{1}{x^8}=(3-2\sqrt{2})^2=17-12\sqrt{2}</math>. Therefore, <math>x^8+\frac{1}{x^8}=24\sqrt{2}+2(17-12\sqrt{2})=34</math>. Also, <math>x^{16}-\frac{1}{x^{16}}=24\sqrt{2}\cdot34=816\sqrt{2}</math>. As a result, the answer of <math>x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}</math> would be <math>816\sqrt{2}+24\sqrt{2}=\boxed{\textbf{(D)} 840\sqrt{2}}</math>.

Latest revision as of 13:42, 26 April 2021

Given that $x^2-\frac{1}{x^2}=2$, what is $x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}$?

$\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\textbf{(C)} ~3780 \qquad\textbf{(D)} ~840\sqrt{2} \qquad\textbf{(E)} ~1260\sqrt{2}$

Solution

We know that $2=x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$. Squaring both sides, we can get that $(x+\frac{1}{x})^2\cdot(x+\frac{1}{x})^2=4$. Therefore, $(x^2+2+\frac{1}{x^2})(x^2-2+\frac{1}{x^2})=4$. Setting $a=x^2+\frac{1}{x^2}$, we can get that $(a+2)(a-2)=4$, and therefore $a^2-4=4$. So $x^2+\frac{1}{x^2}=\sqrt{8}=2\sqrt{2}$. In addition, $x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=2\cdot2\sqrt{2}=4\sqrt{2}$. $\frac{2}{x^2}=2\sqrt{2}-2$, so $\frac{1}{x^2}=\sqrt{2}-1$ and $\frac{1}{x^4}=3-2\sqrt{2}$. Therefore, $x^4+\frac{1}{x^4}=4\sqrt{2}+2(3-2\sqrt{2})=6$. Using a similar approach, we can calculate $x^8-\frac{1}{x^8}=4\sqrt{2}\cdot6=24\sqrt{2}$ and $\frac{1}{x^8}=(3-2\sqrt{2})^2=17-12\sqrt{2}$. Therefore, $x^8+\frac{1}{x^8}=24\sqrt{2}+2(17-12\sqrt{2})=34$. Also, $x^{16}-\frac{1}{x^{16}}=24\sqrt{2}\cdot34=816\sqrt{2}$. As a result, the answer of $x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}$ would be $816\sqrt{2}+24\sqrt{2}=\boxed{\textbf{(D)} 840\sqrt{2}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_April_MIMC_10_Problems/Problem_12&oldid=152732"