How do you simplify #0.5sqrt200 - sqrt(-32)#?

1 Answer
Jul 3, 2017

Answer:

#(5sqrt(8))/2-4sqrt(-2)# or #(5sqrt(8))/2-4isqrt(2)#

Explanation:

#200# can be rewritten as #25x8#, so #sqrt(200)=sqrt(25x8)=sqrt(25)sqrt(8)=5sqrt(8)#

#5sqrt(8)*1/2=(5sqrt(8))/2#

#sqrt(-32)# is impossible to write as a real number so it is taken as #(5sqrt(8))/2-sqrt(-32)#, however #sqrt(x), x<0, = sqrt(abs(x))*i#.

#-32 = -2*16 or 2*-16#. Therefore #sqrt(-32)# is equal to:
#sqrt(-2*16)=sqrt(-2)sqrt(16)=4sqrt(-2)=4*sqrt(2)i=4isqrt(2)#, or
#sqrt(2*-16)=sqrt(2)sqrt(-16)=sqrt(2)4i=4isqrt(2)#

So it can be written as #(5sqrt(8))/2-4isqrt(2)#