Question

$9\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-20=0$

Answer 1

$$\begin{gathered} \text{Given:}9\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-20=0 \\ \mathrm{We}\mathrm{know}\mathrm{the}\mathrm{identity}{x}^2+\frac{1}{x^2}={\left(x-\frac{1}{x}\right)}^2+2 \\ \mathrm{Thus},\text{the given equation can be written as:} \\ 9\left[{\left(x-\frac{1}{x}\right)}^2+2\right]-3\left(x-\frac{1}{x}\right)-20=0 \\ 9{\left(x-\frac{1}{x}\right)}^2+18-3\left(x-\frac{1}{x}\right)-20=0 \\ 9{\left(x-\frac{1}{x}\right)}^2-3\left(x-\frac{1}{x}\right)-2=0 \\ \mathrm{Let}m=x-\frac{1}{x} \\ \mathrm{Then},\text{the equation can be further written as:} \\ 9m^2-3m-2=0 \\ \mathrm{On}\mathrm{splitting}\mathrm{the}\mathrm{middle}\mathrm{term}-3m\mathrm{as}3m-6m,\mathrm{we}\mathrm{get}: \\ 9m^2+3m-6m-2=0 \\ =>3m(3m+1)-2(3m+1)=0 \\ =>(3m+1)(3m-2)=0 \\ =>3m+1=0or3m-2=0 \\ =>m=\frac{-1}{3}orm=\frac{2}{3} \\ \mathrm{On}\mathrm{substituting}m=x-\frac{1}{x},\mathrm{we}\mathrm{get}: \\ x-\frac{1}{x}=\frac{-1}{3}orx-\frac{1}{x}=\frac{2}{3} \\ =>3x^2+x-3=0...\left(1\right)or3x^2-2x-3=0....\left(2\right) \\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ 3x^2+x-3=0 \\ \mathrm{On}\mathrm{using}\mathrm{the}\mathrm{quadratic}\mathrm{formula},\mathrm{we}\mathrm{get} \\ x=\frac{-1\pm \sqrt{(1{)}^2-4(3\left)\right(-3)}}{2\left(3\right)}=\frac{-1\pm \sqrt{1+36}}{6}=\frac{-1\pm \sqrt{37}}{6} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{from}\mathrm{equation}(2),\mathrm{we}\mathrm{get}: \\ 3x^2-2x-3=0 \\ \mathrm{On}\mathrm{using}\mathrm{the}\mathrm{quadratic}\mathrm{formula},\mathrm{we}\mathrm{get}: \\ x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(3\left)\right(-3)}}{2\left(3\right)} \\ =\frac{2\pm \sqrt{4+36}}{6} \\ =\frac{2\pm \sqrt{40}}{6} \\ =\frac{2\pm 2\sqrt{10}}{6}=\frac{1\pm \sqrt{10}}{3} \end{gathered}$$

$\mathrm{Thus},\mathrm{the}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{are}x=\frac{-1\pm \sqrt{37}}{6},\frac{1\pm \sqrt{10}}{3}$