Radical Sums

Relevant wiki: Radical Equations - Basic

$\begin{cases}\sqrt{\color{#D61F06}{a+b}}=25 \ ...(1) \\ \sqrt{a}+\sqrt{b}=31 \ ...(2) \end{cases}$

Squring both sides to $(1)$ .

$\implies \color{#D61F06}{a+b}=(25)^2$

Squring both sides to $(2)$ .

$\implies \underbrace{\color{#D61F06}{a+b}}_{(25)^2}+2\sqrt{\color{#20A900}{ab}}=(31)^2$

$\implies 2\sqrt{\color{#20A900}{ab}}=(31)^2-(25)^2$

$\implies \sqrt{\color{#20A900}{ab}}=\dfrac{56×6}{2}$

$\therefore \sqrt{\color{#20A900}{ab}}=\color{#BA33D6}{\boxed{168}}$ .

By squaring both sides of the equations, we will get:

$\sqrt{a + b} = 25; a + b = 25^2$ $\sqrt{a} + \sqrt{b} = 31; a + 2\sqrt{ab} + b = 31^2$

Subtracting the equations, we will get:

$2\sqrt{ab} = 31^2 - 25^2 = (31 - 25)(31 + 25) = 6\times 56$

Thus, $\sqrt{ab} = 3\times 56 = \boxed{168}$ .