How do you perform the operation and write the result in standard form given #sqrt(-6)*sqrt(-2)#?

1 Answer
George C.
Oct 4, 2016

#sqrt(-6)*sqrt(-2) = -2sqrt(3)#

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 #a < 0# and #b < 0# then:

#sqrt(a) sqrt(b) = -sqrt(ab) != sqrt(ab)#

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