# Question #cabf2

Nov 19, 2016

I found two real solutions:

#### Explanation:

I would try rearranging it as:
${x}^{2} = 4 \sqrt{x}$
square both sides:
${x}^{4} = 16 x$
${x}^{4} - 16 x = 0$
$x \left({x}^{3} - 16\right) = 0$
so we have:
$x = 0$
and
${x}^{3} - 16 = 0$
${x}^{3} = 16$

solutions:
${x}_{1} = 0$
${x}_{2} = \sqrt[3]{16}$

${x}_{3}$ and ${x}_{4}$ should be two imaginary solutions considering that the graph of ${x}^{3} - 16$ has only one real solution ($x$ intercept):
graph{x^3-16 [-105.4, 105.4, -52.7, 52.8]}

Nov 19, 2016

$x = 2.52$

#### Explanation:

${x}^{2} - 4 \sqrt{x} = 0$

${x}^{2} = 4 \sqrt{x}$

${x}^{2} / \sqrt{x} = {x}^{\frac{3}{2}} = 4$

${x}^{\frac{3}{2}} = {\left(\sqrt{x}\right)}^{3} = 4$

$\sqrt{x} = \sqrt[3]{4}$

$x = {\left(\sqrt[3]{4}\right)}^{2} = 2.52$