Question

Find the square root of $\sqrt{48}+\sqrt{45}$.

Answer 1

Given $(\sqrt{48}+\sqrt{45})$

Let $a=(\sqrt{48}+\sqrt{45})$

$=\sqrt{16\times 3}+\sqrt{9\times 5}$

$=4\sqrt{3}+3\sqrt{5}$

$=\sqrt{3}(4+\sqrt{15})$

$=\frac{\sqrt{3}}{2}(8+2\sqrt{3}.\sqrt{5})$

$=\frac{\sqrt{3}}{2}\left\{\right(\sqrt{5}{)}^2+(\sqrt{3}{)}^2+2\sqrt{3}.\sqrt{5}\}$

$a=\left(\frac{\sqrt{3}}{2}\right)(\sqrt{5}+\sqrt{3}{)}^2$.

So square root of $(\sqrt{48}+\sqrt{45})$ is

$\sqrt{(\sqrt{48}+\sqrt{45})}$$=\sqrt{a}=\pm \frac{(\sqrt{3}{)}^{1/2}}{\sqrt{2}}(\sqrt{5}+\sqrt{3})$.