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

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Answer 1 · 2017-10-26T05:42:44

#isqrt(15)#

One identity you need for this is that

#sqrt(a)*sqrt(b)=sqrt(a*b)#

So you can factor out the imaginary number #sqrt(-1)# from the #sqrt(-5)# to get

#sqrt(3)*sqrt(-5) => sqrt(3)*sqrt(5)*sqrt(-1)#

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

#sqrt(3)*sqrt(5)*sqrt(-1) => sqrt(3*5)*sqrt(-1)#

Finally simplify, and replace #sqrt(-1)# with #i#

#sqrt(3)*sqrt(-5) => isqrt(15)#