<s>[INST] 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,
$$x^2-5x+1=0$$
??
 [/INST] For a more brut'ish force alternative, note that the problem assumes $x \ne 0$ for $\sqrt {x} + 1 / \sqrt {x}$ to be defined, so that $x$ must be a root of $x^2-5x+1=0 \iff x = \frac{1}{2}\left(5 \pm \sqrt{21}\right)\,$.
By known radical denesting techniques, $\sqrt{5 \pm \sqrt{21}}=\frac{1}{2}\left(\sqrt{14} \pm \sqrt{6}\right)\,$, so:
$$\require{cancel}
\sqrt{x}=\frac{1}{2 \sqrt{2}}\left(\sqrt{14} \pm \sqrt{6}\right) = \frac{1}{2}\left(\sqrt{7}\pm\sqrt{3}\right) \;\;\implies\;\; \frac{1}{\sqrt{x}} = \frac{1}{2}\left(\sqrt{7} \mp \sqrt{3}\right)
$$
Then $\displaystyle x + \frac{1}{\sqrt{x}}=\frac{1}{2}\left(\sqrt{7}\pm\cancel{\sqrt{3}}\right) + \frac{1}{2}\left(\sqrt{7} \mp \cancel{\sqrt{3}}\right)=2 \cdot \frac{1}{2} \sqrt{7} = \sqrt{7}$.
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