How do I solve #sqrt(-8)(sqrt(-3)-sqrt(5))#?

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Answer 1 · 2016-11-21T04:52:23

#-2sqrt(6)-(2sqrt10)i#
or
#-2sqrt2(sqrt3+isqrt5)#

"Simplify" is perhaps a better word than "solve", since there's no variable to solve for. First, we rewrite using #i# notation:

#sqrt(-8)(sqrt(-3)-sqrt5)#

#=isqrt8(isqrt3-sqrt5)#

#=2isqrt2(isqrt3-sqrt5)#

Next, distribute:

#=2i^2sqrt6-2isqrt10#

Then, we remember that #i^2=-1# by definition, so we can write

#=-2sqrt(6)-(2sqrt10)i#

This is now in #x+yi# form. It can be "simplified" further, however, by factoring out -#2sqrt2# from both terms:

#=-2sqrt2(sqrt3+isqrt5)#