• # question_answer $\sqrt{-2}\,\sqrt{-3}=$ [Roorkee 1978] A) $\sqrt{6}$ B) $-\sqrt{6}$ C) $i\sqrt{6}$ D) None of these

$\sqrt{-2}\sqrt{-3}=i\sqrt{2}\,i\,\sqrt{3}={{i}^{2}}\sqrt{6}=-\sqrt{6}$