How do you simplify #(2sqrt50) / sqrt2#?

1 Answer
Pythagoras
Sep 20, 2015

10

Explanation:

Oh I like this one.

Well, you have to notice that #2=sqrt(2)*sqrt(2)#
and rewrite as such:

#(sqrt2 *sqrt2*sqrt50)/(sqrt2)#

You can then cancel out a #sqrt2# both top and bottom and get
#=sqrt2*sqrt50#

Which is easy because the multiplication of square-roots is like the square-root of the multiplication, I mean:
#sqrt(a)*sqrt(b)=sqrt(a*b)#
So we have:
#=sqrt(2*50)#
#=sqrt(100)#
#=10#

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