Question

Solve the following equations:
$9+x^{-4}=10x^{-2}$.

Answer 1

$9+x^{-4}=10x^{-2}9+\frac{1}{x^4}=\frac{10}{x^2}$

Put $\frac{1}{x^2}=t$

$9+{\left(\frac{1}{x^2}\right)}^2=10\frac{1}{x^2}9+t^2=10t{t}^2-10t+9=0t^2-t-9t+9=0t(t-1)-9(t-1)=0(t-9)(t-1)=0t=1,9x^2=\frac{1}{t}{x}^2=1,\frac{1}{9}x=\sqrt{1},\sqrt{\frac{1}{9}}x=\pm 1,\pm \frac{1}{3}$