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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 4"

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==Solution==
 
==Solution==
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asdf
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From the problem, we know that
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<cmath>\frac{x+2}{6} = \frac{6}{x+2}</cmath>
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<cmath>(x+2)^2 = 6^2</cmath>
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<cmath>x^2+ 4x + 4 = 36</cmath>
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<cmath>x^2 + 4x - 32 = 0</cmath>
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<cmath>(x-8)(x+4) = 0</cmath>
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Thus, <math>x = 8</math> or <math>x = -4</math>. Our answer is <math>8 \cdot(-4)=\boxed{-32}</math>
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~Bradygho

Revision as of 21:09, 10 July 2021

Problem

If $\frac{x+2}{6}$ is its own reciprocal, find the product of all possible values of $x.$

Solution

From the problem, we know that \[\frac{x+2}{6} = \frac{6}{x+2}\] \[(x+2)^2 = 6^2\] \[x^2+ 4x + 4 = 36\] \[x^2 + 4x - 32 = 0\] \[(x-8)(x+4) = 0\]

Thus, $x = 8$ or $x = -4$. Our answer is $8 \cdot(-4)=\boxed{-32}$

~Bradygho

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