Difference between revisions of "2021 April MIMC 10 Problems/Problem 10"
Cellsecret (talk | contribs) (Created page with "If <math>x+\frac{1}{x}=-2</math> and <math>y=\frac{1}{x^{2}}</math>, find <math>\frac{1}{x^{4}}+\frac{1}{y^{4}}</math>. <math>\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\t...") |
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<math>\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\textbf{(C)} ~0 \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2</math> | <math>\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\textbf{(C)} ~0 \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2</math> | ||
==Solution== | ==Solution== | ||
| − | To | + | The simplest way to solve this problem is to solve <math>x</math> and <math>y</math>, respectively. To solve <math>x</math>, we can multiply both sides of <math>x+\frac{1}{x}=-2</math> by <math>x</math>, and this gives us the quadratic <math>x^2+2x+1=0</math>. Solve the quadratic, <math>x=-1</math>. Substitute <math>x</math> into the second equation to evaluate <math>y</math>, <math>y=1</math>. <math>\frac{1}{1}+\frac{1}{1}=\fbox{\textbf{(E)} 2}</math>. |
Latest revision as of 12:39, 26 April 2021
If 
and 
, find 
.
Solution
The simplest way to solve this problem is to solve 
and 
, respectively. To solve , we can multiply both sides of by , and this gives us the quadratic 
. Solve the quadratic, 
. Substitute into the second equation to evaluate , 
. 
.
