How do you perform the operation and write the result in standard form given #sqrt(-6)*sqrt(-2)#?
1 Answer
Oct 4, 2016
Explanation:
Be a little careful!
#sqrt(-6) * sqrt(-2) = (sqrt(6)i) * (sqrt(2)i)#
#color(white)(sqrt(-6)*sqrt(-2)) = sqrt(6)sqrt(2)*i^2#
#color(white)(sqrt(-6)*sqrt(-2)) = -sqrt(6*2)#
#color(white)(sqrt(-6)*sqrt(-2)) = -sqrt(2^2*3)#
#color(white)(sqrt(-6)*sqrt(-2)) = -2sqrt(3)#
Note that if
#sqrt(a) sqrt(b) = -sqrt(ab) != sqrt(ab)#