View source for 2021 JMPSC Accuracy Problems/Problem 4

==Problem==
If <math>\frac{x+2}{6}</math> is its own reciprocal, find the product of all possible values of <math>x.</math>

==Solution==
From the problem, we know that 
<cmath>\frac{x+2}{6} = \frac{6}{x+2}</cmath>
<cmath>(x+2)^2 = 6^2</cmath>
<cmath>x^2+ 4x + 4 = 36</cmath>
<cmath>x^2 + 4x - 32 = 0</cmath>
<cmath>(x+8)(x-4) = 0</cmath>

Thus, <math>x = -88</math> or <math>x = 4</math>. Our answer is <math>(-8) \cdot 4=\boxed{-32}</math>

~Bradygho

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

~Geometry285

==Solution 3==
<math>\frac{x+2}{6}=\frac{6}{x+2} \implies x^2+4x-32</math> Therefore, the product of the root is <math>-32</math> 

~kante314

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

~tigerzhang