2021 JMPSC Accuracy Problems/Problem 4

Problem

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

~Bradygho

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

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

~kante314

The only numbers that are their own reciprocals are $1$ and $-1$ . Thus,