How do you simplify #sqrt6 - sqrt 3#?

1 Answer
May 29, 2016

Answer:

#sqrt6-sqrt3=sqrt3(sqrt2-1)#

Explanation:

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

Recall that (for #a,b>0#)

#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 #sqrt3# from both terms:

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

I wouldn't go so far as to say that #sqrt3(sqrt2-1)# is a simplification of #sqrt6-sqrt3#, but they are equivalent statements. Using #sqrt3(sqrt2-1)# may be helpful in simplifying more complicated expressions, such as:

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