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

$$\left(\frac{x+1}{\sqrt{x}}\right)^2 = \frac{(x^2+2x+1)}x  =  \frac{(x^2-5x+1)+7x}x  =  7x/x = 7$$($x$ is not equal to $0$)

Therefore, $$\sqrt{x}+\frac 1{\sqrt{x}} = \sqrt{7}$$