How do you simplify #sqrt0.08#?

1 Answer
Mar 16, 2016

You may begin by writing #sqrt0.08=sqrt(8/100)#

Explanation:

Then we apply the 'radical rules':
#sqrt(a/b)=sqrta/sqrtb# and #sqrt(axxb)=sqrtaxxsqrtb#

Translated to our problem:
#=sqrt8/sqrt100=sqrt8/sqrt(10^2)=sqrt8/10#

But since #8=4xx2=2^2xx2#:
#->sqrt8=sqrt(2^2)xxsqrt2=2sqrt2#

Then we get to:
#=(2sqrt2)/10=1/5sqrt2#