Where does the fourth step come from?

Question

I'm talking about the step after $x(x-1) = \sqrt\frac{\mu}{3}$.

Answer 1

You have $$x(x-1)=\sqrt{\frac{\mu}{3}}$$

$$x^2-x-\sqrt{\frac{\mu}{3}}=0$$

Now completing the square gives

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

so $$x=\frac{1}{2}\pm\sqrt{\sqrt{\mu/3}+\frac{1}{4}}$$

Then since you want the solution with $x>1$, you take the positive root and hence $$x=\frac{1}{2}+\sqrt{\sqrt{\mu/3}+\frac{1}{4}}.$$

Answer 2

This step follows after the resolution of the quadratic equation

$$x(x-1) = \sqrt\frac{\mu}{3} \iff x^2-x-\sqrt\frac{\mu}{3}=0$$

with the condition $x\ge1$.