​ Prove that √{-16}+3√{-25}+√{-36}-√{-625} = 0. ​

Question

Prove that \(\sqrt{-16}+3\sqrt{-25}\) \(+\sqrt{-36}-\sqrt{-625}\) = 0.

Answer 1

L.H.S = \(\sqrt{-16}+3\sqrt{-25}\) \(+\sqrt{-36}-\sqrt{-625}\)

Since we know that i = \(\sqrt{-1}\).

So,

  \(\sqrt{16}_{i+3}\sqrt{25}i\) \(+\sqrt{36}i-\sqrt{625}_i\)

= 4i + 15i + 6i - 25i

= 0

L.H.S = R.H.S

Hence proved.