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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