How do you divide #(sqrt6+sqrt3)/ sqrt2#?

Question

1362 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-16T16:25:11

#sqrt(3)+sqrt(6)/2#

Separate the numerator using the common denominator of #sqrt(2)#

#sqrt(6)/sqrt(2)+sqrt(3)/sqrt(2)#

Write using single radicals, #sqrt()#

#sqrt(6/2)+sqrt(3)/sqrt(2)#

Simplify the first term

#sqrt(3)+sqrt(3)/sqrt(2)#

Rationalize the second term by multiplying the numerator and denominator by #sqrt(2)#

#sqrt(3)+sqrt(3)/sqrt(2)*sqrt(2)/sqrt(2)=sqrt(3)+sqrt(6)/2#