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1953 AHSME Problems/Problem 8

Revision as of 10:59, 7 June 2016 by Katzrockso (talk | contribs) (Created page with "==Problem 8== The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is: <math>\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B...")
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Problem 8

The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:

$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$

Solution

$x+y=2\implies y=2-x$. Then $2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8$. We now notice that $x=0\implies (2)(4)=8$, so $\fbox{C}$

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