<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] If $y=\sqrt x + 1/\sqrt x$ then your value of $y$ is given as a solution of $y$ for the system
\begin{align}
x^2&-5x+1=0\\
t^2&=x\\
t^2&=ty-1\\
\end{align}
Putting the second and third equation together $x=ty-1$ so
$$(ty-1)^2-5(ty-1)+1=t^2y^2-2ty+1-5ty+5+1=(ty-1)y^2-7(ty-1)=0$$
If $ty-1\neq 0$ then $y^2=7$.
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