# How do you simplify #sqrt6 - sqrt 3#?

##### 1 Answer

May 29, 2016

#### Explanation:

This is already about as simplified as it could be. However, if you wanted to, this *can* be factored.

Recall that (for

#sqrt(ab)=sqrta*sqrtb#

Therefore we can say that

#sqrt6=sqrt(2*3)=sqrt2*sqrt3#

We can then rewrite the original expression:

#sqrt6-sqrt3=(sqrt2*sqrt3)-sqrt3#

Factor

#(sqrt2*sqrt3)-sqrt3=sqrt3(sqrt2-1)#

I wouldn't go so far as to say that

#(sqrt6-sqrt3)/(sqrt10-sqrt5)=(sqrt3(sqrt2-1))/(sqrt5(sqrt2-1))=sqrt3/sqrt5=sqrt3/sqrt5(sqrt5/sqrt5)=sqrt15/5#