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revised Area of a triangle from some of its parts
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28
comment Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$?
@SufyanNaeem: No. First off, $b\neq 0$ since we're dividing by $b$. Beyond that, your example "since..." is not true because of the broad rule $\sqrt{ab}=\sqrt{a}\sqrt{b}$, but because of the narrower rule $\sqrt{-a}=i\sqrt{a}$ for $a>0$. In particular, take another look at the paragraph beginning "In general..." in my answer.
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