Question

Prove that:

$3\sqrt{48}-4\sqrt{75}+\sqrt{192}=0$

Answer 1

We have,

$3\sqrt{48}-4\sqrt{75}+\sqrt{192}$

$=3\sqrt{16\times 3}-4\sqrt{25\times 3}+\sqrt{64\times 3}$

$=12\sqrt{3}-20\sqrt{3}+8\sqrt{3}$

$=20\sqrt{3}-20\sqrt{3}$

$=0$