How do you simplify #((sqrt 2) + 2 (sqrt 2) + (sqrt8)) / (sqrt 3)#?

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Answer 1 · 2016-05-04T12:47:11

#(5sqrt(6))/3#

Consider #sqrt(8) -> sqrt(2xx2^2)=2sqrt(2)#

Write the given expression as:

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

#(5sqrt(2))/sqrt(3)#

But it is not 'good form' to have a root in the denominator. So we need 'get rid' of it if we can.

Multiply by 1 but in the form of #1=sqrt(3)/sqrt(3)#

#(5sqrt(2))/sqrt(3)xxsqrt(3)/sqrt(3) = (5sqrt(6))/3#

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Note that #sqrt(2)xxsqrt(3)=sqrt(2xx3)=sqrt(6)#