How do you add #(3sqrt3 + 9sqrt3 + sqrt6 + 3sqrt9) /(sqrt4 - 9sqrt9)#?

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Answer 1 · 2015-04-13T10:40:36

The first thing you may notice, is that the numerator is easy to simplify:

#"numerator"=sqrt4-9sqrt9=2-9*3=-25#

We will forget that for a while and focus on the denominator:

Add like roots and simplify #sqrt9=3#

#"denominator"=(3+9)sqrt3+sqrt6+3*3#

Since you can't simplify this any further:

#"answer"=-(12sqrt3+sqrt6+9)/25#