Question

Solve : ${9x}^2+\frac{1}{{9x}^2}-2=0$

Answer 1

Given eqaution is $9x^2+\frac{1}{9x^2}-2=0$

$\Rightarrow (3x{)}^2+\frac{1}{(3x{)}^2}+2.3x\times \frac{1}{3x}-2\times 3x\times \frac{1}{3x}-2=0$

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

$\Rightarrow (3x+\frac{1}{3x}{)}^2=4$

$\Rightarrow 3x+\frac{1}{3}x=\pm 2$

$\Rightarrow 9x^2\pm 6x+1=0$

$\Rightarrow (3x{)}^2\pm 2.3x.1+1^2=0$

$\Rightarrow (3x+1{)}^2=0$ $\therefore x=\pm \frac{1}{3}$