How do you simplify # 3sqrt(-8) • isqrt2#?

1 Answer
George C.
Dec 20, 2015

#3sqrt(-8)*isqrt(2) = -12#

Explanation:

Any non-zero number has two square roots, but when we write #sqrt(x)# we want it to indicate one of the roots unambiguously, which is called the principal square root.

If #x > 0# then both of its square roots are Real. By definition the principal square root #sqrt(x)# is the positive one.

If #x < 0# then both of its square roots are pure imaginary. By definition the principal square root #sqrt(x)# is the one with a positive coefficient of #i#, that is #i sqrt(-x)#.

If #x < 0# then #sqrt(x) = i sqrt(-x)#

With these definitions of #sqrt(x)#, the following identity holds when #a, b >= 0#:

If #a, b >= 0# then #sqrt(a)sqrt(b) = sqrt(ab)#

Note carefully that #sqrt(a)sqrt(b) = sqrt(ab)# does not work if #a, b < 0#. For example:

#1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = i sqrt(1) * i sqrt(1) = i^2 = -1#

With this in mind, we exercise some care to write:

#3sqrt(-8)*isqrt(2)#

#=3isqrt(8)*isqrt(2)#

#=3i^2 sqrt(8*2)#

#= -3sqrt(16)#

#=-3*4=-12#

Related questions
See all questions in Multiplication of Complex Numbers
Impact of this question
1668 views around the world
You can reuse this answer
Creative Commons License