Question

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

Answer 1

Now,

$\sqrt{48}-\sqrt{45}$

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

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

$=\frac{\sqrt{3}}{2}(8-2\sqrt{15})$

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

$=\frac{\sqrt{3}}{2}\{\sqrt{5}-\sqrt{3}{\}}^2$

Now square root of the given number is $\pm \frac{3^{\frac{1}{4}}}{\sqrt{2}}(\sqrt{5}-\sqrt{3})$.