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,
$$x^2-5x+1=0$$
??
 A: $x^3-5x^2+x$ 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 $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$
A: \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*}
A: 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 $y^2=7$.
A: 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 $\displaystyle x + \frac{1}{\sqrt{x}}=\frac{1}{2}\left(\sqrt{7}\pm\cancel{\sqrt{3}}\right) + \frac{1}{2}\left(\sqrt{7} \mp \cancel{\sqrt{3}}\right)=2 \cdot \frac{1}{2} \sqrt{7} = \sqrt{7}$.
A: $$\sqrt{x}+\frac 1{\sqrt{x}} = \frac{(x+1)}{\sqrt{x}}$$  
$$\left(\frac{x+1}{\sqrt{x}}\right)^2 = \frac{(x^2+2x+1)}x  =  \frac{(x^2-5x+1)+7x}x  =  7x/x = 7$$($x$ is not equal to $0$)
Therefore, $$\sqrt{x}+\frac 1{\sqrt{x}} = \sqrt{7}$$
