If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$

Question

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

Answer 1

\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*}

Answer 2

$x^3-5x^2+x$ 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 $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$

Answer 3

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 $y^2=7$.

Answer 4

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 $\displaystyle x + \frac{1}{\sqrt{x}}=\frac{1}{2}\left(\sqrt{7}\pm\cancel{\sqrt{3}}\right) + \frac{1}{2}\left(\sqrt{7} \mp \cancel{\sqrt{3}}\right)=2 \cdot \frac{1}{2} \sqrt{7} = \sqrt{7}$.

Answer 5

$$\sqrt{x}+\frac 1{\sqrt{x}} = \frac{(x+1)}{\sqrt{x}}$$

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

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