How to solve for $x$ in ${\sqrt{9+2x}} - {\sqrt{2x}} = \frac{5}{\sqrt{9+2x}}$

Question

How can I solve for $x$ in the following equation?

${\sqrt{9+2x}} - {\sqrt{2x}} = \frac{5}{\sqrt{9+2x}}$

Answer 1

HINT:

Clearly, $9+2x\ne0$

$\implies9+2x-\sqrt{2x(9+2x)}=5\iff\sqrt{2x(9+2x)}=9+2x-5$

Squaring we get $2x(9+2x)=(4+2x)^2$

Can you proceed from here?

Test for the validity of the root

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$(\sqrt{9+2x}+\sqrt{2x})(\sqrt{9+2x}-\sqrt{2x})=9$

$$\sqrt{9+2x}-\sqrt{2x}=\dfrac5{\sqrt{9+2x}}$$

$$\iff\sqrt{9+2x}+\sqrt{2x}=\dfrac{9\sqrt{9+2x}}5$$

$$\iff\sqrt{2x}=\dfrac45\sqrt{9+2x}$$