Prove that for every real number $x$, if $x>2$ then there is a real number $y$ such that $y+1/y = x$. I'm trying to learn about proofs and I'm stuck in Velleman's book "How to Prove it".

This is the question (ex.6 p.118):


  Prove that for every real number $x$, if $x &gt; 2$ then there is a real number $y$ such that $y+\frac{1}{y}= x$.


I couldn't do it so I went to see the answer, and it is written that we can assume that $y=(x+\sqrt{x^2-4})/2$.

How did the author get to this value for $y$?
Suppose that

$$y+\frac{1}{y}=x.$$

Then

$$y^2+1=xy,$$

and

$$y^2-xy+1=0.$$

It follows that

$$y=\frac{x\pm\sqrt{x^2-4}}{2}$$

either by completing the square, or by using the quadratic formula.
The author just pulled it out of nowhere - where it comes from isn't important to the proof, only that it works. For how they knew what to pull out of nowhere, try rearranging $y + \frac{1}{y} = x$ (which I suspect is what the formula was supposed to read?) to make $x$ the subject.