2021 JMPSC Accuracy Problems/Problem 4

Problem

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

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, $x = -88$ 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 $4 \cdot (-8)=\boxed{-32}$ .

Page

Toolbox

Search