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2021 JMPSC Accuracy Problems/Problem 4

Revision as of 16:23, 11 July 2021 by Samrocksnature (talk | contribs)
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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

Solution 2

We have $\frac{x+2}{6} = \frac{6}{x+2}$, so $x^2+4x-32=0$. By Vieta's our roots $a$ and $b$ amount to $\frac{-32}{1}=\boxed{-32}$

~Geometry285

Solution 3

$\frac{x+2}{6}=\frac{6}{x+2} \implies x^2+4x-32$ Therefore, the product of the root is $-32$

~kante314

Solution 4

The only numbers that are their own reciprocals are $1$ and $-1$. The equation $\frac{x+2}{6}=1$ has the solution $x=4$, while the equation $\frac{x+2}{6}=-1$ has the solution $x=-8$. The answer is $4 \cdot (-8)=\boxed{-32}$.

~tigerzhang


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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