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2021 JMPSC Accuracy Problems/Problem 4

Revision as of 13:20, 11 July 2021 by Bradygho (talk | contribs) (→‎Solution)

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

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