Question

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

Answer 1

$$\begin{gathered} \text{Given:}3\left(x^2+\frac{1}{x^2}\right)-4\left(x-\frac{1}{x}\right)-6=0 \\ \mathrm{We}\mathrm{know}\mathrm{the}\mathrm{identity}{x}^2+\frac{1}{x^2}={\left(x-\frac{1}{x}\right)}^2+2 \\ \mathrm{Thus},\text{the equation can be written as}: \\ 3\left[{\left(x-\frac{1}{x}\right)}^2+2\right]-4\left(x-\frac{1}{x}\right)-6=0 \\ 3{\left(x-\frac{1}{x}\right)}^2+6-4\left(x-\frac{1}{x}\right)-6=0 \\ 3{\left(x-\frac{1}{x}\right)}^2-4\left(x-\frac{1}{x}\right)=0 \\ \mathrm{Let}m=x-\frac{1}{x} \\ \mathrm{Then},\text{the equation can be further written as:} \\ 3m^2-4m=0 \\ m(3m-4)=0 \\ =>m=0or3m-4=0 \\ =>m=0orm=\frac{4}{3} \end{gathered}$$
$\text{On}\mathrm{substituting}m=x-\frac{1}{x},\mathrm{we}\mathrm{get}:x-\frac{1}{x}=0orx-\frac{1}{x}=\frac{4}{3}$

$$\begin{gathered} \Rightarrow \frac{x^2-1}{x}=0or\frac{x^2-1}{x}=\frac{4}{3} \\ \Rightarrow x^2=1or3\left(x^2-1\right)=4x \\ \Rightarrow x=\pm 1or3x^2-3-4x=0 \\ \Rightarrow x=\pm 1or3x^2-4x-3=0 \\ \Rightarrow x=\pm 1orx=\frac{4\pm \sqrt{{\left(4\right)}^2-4\times 3\times \left(-3\right)}}{2\times 3} \\ \Rightarrow x=\pm 1orx=\frac{4\pm \sqrt{16+36}}{6} \\ \Rightarrow x=\pm 1orx=\frac{4\pm \sqrt{52}}{6} \\ \Rightarrow x=\pm 1orx=\frac{4\pm \sqrt{52}}{6} \\ \Rightarrow x=\pm 1orx=\frac{2\pm \sqrt{13}}{3} \end{gathered}$$