An equation linear in the square root

Statement

Solve the equation $3\sqrt{x} + 5 = 20 - 2\sqrt{x}$.

Solution

Existence condition: $x \ge 0$.

Treat $\sqrt{x}$ as a single quantity. Move the terms with $\sqrt{x}$ to the left: $3\sqrt{x} + 2\sqrt{x} = 20 - 5 \implies 5\sqrt{x} = 15 \implies \sqrt{x} = 3.$ Squaring: $x = 9$ (acceptable since $\ge 0$). Check: $3\cdot 3 + 5 = 14$ and $20 - 2\cdot 3 = 14$. Correct. $\boxed{x = 9}$