<s>[INST] Prove that if $x \in \mathbb R$ and $x>2$ then $y+ \frac{1}{y} = x$ will have real solution. 
Prove that for real number $x$, if $x > 2$ then there is real number $y$ such that $y + \frac{1}{y} = x$

My attempt: 
Rewriting equation, we have:
$$\tag1y + \frac{1}{y} = x$$
$$\tag2 \frac{y^2+ 1}{y} = x$$
$$\tag3 y^2+ 1 = yx$$
$$\tag4 y^2 - yx + 1 = 0$$
We know that for quadratic equations of the format $aq^2 + bq + c$ ($a,b,c$ are constants and q is variable), if discriminant is positive, then equation will have real solutions.
Let discriminant be denoted by $D$, then
$D = \sqrt{b^2- 4ac}$
Substituting our values into the equation, we have:
$D = \sqrt{x^2 - 4}$
It can be seen that if $x > 2$ then $D$ is positive, hence $y + \frac{1}{y} = x$ will have real solution.
Is it correct?
 [/INST] Multiplying both sides by $y$ and putting the result in standard form, we have
$$y^{2} - xy + 1 = 0.$$
Hence, the discriminate of this quadratic equation is $x^{2} - 4$; and so, if $x > 2$, we certainly get two distinct real solutions. But also, if $x < -2$ we may draw the same conclusion as well. Finally, if $x = \pm 2$, then we get exactly one real solution.
</s>