$\frac{1}{x^2}\ = (-5+\sqrt{3})$

Question

How can I solve the equation: $$\frac{1}{x^2}\ = (-5+\sqrt{3})$$ I tried this: $$ x^2\cdot\frac{1}{x^2} = x^2\cdot(-5+\sqrt{3}) $$ $$ 1=-5x^2+\sqrt{3}x^2 $$ $$ 1= x^2(-5+\sqrt{3}) $$ $$ x^2=\frac{1}{(-5+\sqrt{3})} $$ $$ x=\pm \sqrt{\frac{1}{-5+\sqrt{3}}} $$ Not sure if this is correct..

Answer 1

Your answer is correct. Roots are imaginary as follows $$\frac1{x^2}=-5+\sqrt3$$ $$x^2=\frac{1}{-5+\sqrt3}$$ $$x^2=\frac{-5-\sqrt3}{(-5+\sqrt3)(-5-\sqrt3)}$$

$$x^2=\frac{-5-\sqrt3}{22}$$

$$x=\pm i\sqrt{\frac{5+\sqrt3}{22}}$$

Answer 2

This equation has no real roots because $-5+\sqrt3<0$

In $\mathbb C$ we obtain: $$\left\{\frac{1}{\sqrt{5-\sqrt3}}i,-\frac{1}{\sqrt{5-\sqrt3}}i\right\}$$