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

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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==
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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>
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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 $x+\frac{1}{x}=-2$ and $y=\frac{1}{x^{2}}$, find $\frac{1}{x^{4}}+\frac{1}{y^{4}}$.

$\textbf{(A)} ~-2 \qquad\textbf{(B)} ~-1 \qquad\textbf{(C)} ~0 \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution

The simplest way to solve this problem is to solve $x$ and $y$, respectively. To solve $x$, we can multiply both sides of $x+\frac{1}{x}=-2$ by $x$, and this gives us the quadratic $x^2+2x+1=0$. Solve the quadratic, $x=-1$. Substitute $x$ into the second equation to evaluate $y$, $y=1$. $\frac{1}{1}+\frac{1}{1}=\fbox{\textbf{(E)} 2}$.

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