2021 JMPSC Accuracy Problems/Problem 4


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


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 $(-8) \cdot 4=\boxed{-32}$


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 $\frac{-32}{1}=\boxed{-32}$


Solution 3

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


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 $4 \cdot (-8)=\boxed{-32}$.


See also

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

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