If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$ 
And,
$$x^2-5x+1=0$$
??

$x^3-5x^2+x$ gives $x=0$ or $x^2+1=5x$.
For $x\leq0$ the needed value does not exist.
For $x>0$ we have $\sqrt{x}+\frac{1}{\sqrt{x}}>0$.
Thus,
$$x+\frac{1}{x}=5$$
or $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=7,$$
which gives $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$
