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,
$$x^2-5x+1=0$$
??

\begin{align*}
x^3+x&=5x^2\\
x+\frac{1}{x}&=5\\
\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2&=7\\
\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)&=\sqrt{7}.
\end{align*}
