Prove that

Question:

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

 

Solution:

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

Since we know that $\mathrm{i}=\sqrt{-1}$.

So,

$=\sqrt{16} i+3 \sqrt{25} i+\sqrt{36} i-\sqrt{625} i$

$=4 \mathrm{i}+15 \mathrm{i}+6 \mathrm{i}-25 \mathrm{i}$

$=0$

L.H.S = R.H.S

Hence proved.

 

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