Question

Solve the following equation.

$3(x^2+\frac{1}{x^2})-4(x-\frac{1}{x})-6=0$

• A. $x=1,-1,\frac{-2\pm \sqrt{13}}{3}$
• B. $x=1,-1,\frac{2\pm \sqrt{13}}{3}$
• C. $x=1,-1,\frac{4\pm \sqrt{13}}{6}$
• D. None of these

Answer 1

The correct option is B $x=1,-1,\frac{2\pm \sqrt{13}}{3}$

$3(x^2+\frac{1}{x^2})-4(x-\frac{1}{x})-6=0$
Let $(x-\frac{1}{x})=m$
then $(x-\frac{1}{x}{)}^2=m^2$
$x^2+\frac{1}{x^2}-2=m^2$
$x^2+\frac{1}{x^2}=m^2+2$
Hence, reframing the equation,
$3(m^2+2)-4m-6=0$
$3m^2-4m=0$
$m=0,\frac{4}{3}$
Now, when m = 0, $x-\frac{1}{x}=0$
$x^2-1=0$ or $x=\pm 1$
Now, when $m=\frac{4}{3}$, $x-\frac{1}{x}=\frac{4}{3}$
$3x^2-3=4x$
$3x^2-4x-3=0$
$x=\frac{4\pm \sqrt{52}}{6}$ = $\frac{2\pm \sqrt{13}}{3}$