Can you tell me what is wrong of my solution that is not found in possible answers which are $1, {{5}{/}{2}}, \sqrt{2}, 2, 3 $
$$ \text { if } \quad x+\frac{2}{\sqrt{x}}=5 \quad \text {, what is} \quad x+2 \sqrt{x}= \text { ?} $$
My approach is the following but I couldn't get rid of $x$
$x\sqrt{x}+2\:=\:5\sqrt{x}$
$\left(5-x\right)\sqrt{x}=2$
$5-x=\frac{2\sqrt{x}}{x}$
$5x-x^2=2\sqrt{x}$
$\color{blue}{-6x} +5x-x^{2} =\color{blue}{-6x} +2\sqrt{x}$
$x^2+x=6x-2\sqrt{x}$
$x+2\sqrt{x}=6x-x^2$
$$x + \frac{2}{\sqrt{x}} = 5$$ Let $t = \sqrt{x}$. Then $t^2 = x$ and $x \geq 0$. Therefore, we obtain \begin{align*} t^2 + \frac{2}{t} & = 5\\ t^3 + 2 & = 5t\\ t^3 - 5t + 2 & = 0 \end{align*} By the Rational Roots Theorem, the possible rational roots are $\pm 1, \pm 2$.
Since $t = \sqrt{x} \geq 0$, the only possible rational roots are $1, 2$. Substitution shows that $t = 2$ is a root, while $t = 1$ is not a root. Hence, $t - 2$ is a factor. We can find the remaining factor by factoring or dividing. \begin{align*} t^3 - 5t + 2 & = t^2(t- 2) + 2t^2 - 5t + 2\\ & = t^2(t - 2) + 2t(t - 2) + 4t - 5t + 2\\ & = t^2(t - 2) + 2t(t - 2) - t + 2\\ & = t^2(t - 2) + 2t(t - 2) - 1(t - 2)\\ & = (t^2 + 2t - 1)(t - 2) \end{align*} Hence, $$(t^2 + 2t - 1)(t - 2) = 0$$ Since a product is equal to zero if and only if one of its factors is equal to zero, \begin{align*} t^2 + 2t - 1 & = 0 & \text{or} & & t - 2 & = 0\\ t^2 + 2t & = 1 & & & t & = 2\\ t^2 + 2t + 1 & = 2\\ (t + 1)^2 & = 2\\ |t + 1| & = \sqrt{2}\\ \end{align*} If $|t + 1| = \sqrt{2}$, then \begin{align*} t + 1 & = \sqrt{2} & \text{or} & & t + 1 & = -\sqrt{2}\\ t & = -1 + \sqrt{2} & & & t & = -1 - \sqrt{2} \end{align*} Since $t = \sqrt{x} \geq 0$, we may discard the solution $t = -1 - \sqrt{2}$.
Since $x = t^2$, the solutions of the equation are $x = 4$ and $x = 3 - 2\sqrt{2}$.
You can substitute those values into the expression $x + 2\sqrt{x}$ to complete the question.