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Answer 1 · 2018-04-09T23:29:34

$\pm 5 i$ $+-5i$

Remember: $\sqrt{-} 1 = i$ $sqrt-1=i$ and $\sqrt{a b} = \sqrt{a \cdot b}$ $sqrt(ab)=sqrt(a*b)$

25 is a perfect square ( $5^{2} = 25$ $5^2=25$ , ${(- 5)}^{2} = 25$ $(-5)^2=25$ ), separate that from -1

$\pm \sqrt{25 \cdot - 1} \to \sqrt{-} 1$ $+-sqrt(25*-1) rarr sqrt-1$ is $i$ $i$

$\pm 5 i \to$ $+- 5i rarr$ The $\pm$ $+-$ symbol shows that the answer can be negative or positive

Answer 2 · 2018-04-09T23:31:48

$\pm 5 i$ $+-5i$

Remember $i = \sqrt{- 1}$ $i=sqrt(-1$

So:
$\pm \sqrt{- 25} =$ $+-sqrt(-25)=$

$\pm \sqrt{- 1 \cdot 25} =$ $+-sqrt(-1*25)=$

$\pm i \sqrt{25} =$ $+-isqrt(25)=$

$\pm 5 i$ $+-5i$