How do you evaluate $\sqrt{3} \cdot \sqrt{- 5}$ $\sqrt { 3} \cdot \sqrt { - 5}$ ?

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Answer 1 · 2017-10-26T05:42:44

$i \sqrt{15}$ $isqrt(15)$

One identity you need for this is that

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$ $sqrt(a)*sqrt(b)=sqrt(a*b)$

So you can factor out the imaginary number $\sqrt{- 1}$ $sqrt(-1)$ from the $\sqrt{- 5}$ $sqrt(-5)$ to get

$\sqrt{3} \cdot \sqrt{- 5} \Rightarrow \sqrt{3} \cdot \sqrt{5} \cdot \sqrt{- 1}$ $sqrt(3)*sqrt(-5) => sqrt(3)*sqrt(5)*sqrt(-1)$

Then, we use the identity again to combine the real number parts

$\sqrt{3} \cdot \sqrt{5} \cdot \sqrt{- 1} \Rightarrow \sqrt{3 \cdot 5} \cdot \sqrt{- 1}$ $sqrt(3)*sqrt(5)*sqrt(-1) => sqrt(3*5)*sqrt(-1)$

Finally simplify, and replace $\sqrt{- 1}$ $sqrt(-1)$ with $i$ $i$

$\sqrt{3} \cdot \sqrt{- 5} \Rightarrow i \sqrt{15}$ $sqrt(3)*sqrt(-5) => isqrt(15)$

How do you evaluate √3⋅√−5\sqrt { 3} \cdot \sqrt { - 5}?