How do you simplify: #(sqrt2+ 2sqrt2 +sqrt8) /sqrt3#?

2 Answers
Aug 13, 2018

Answer:

#(sqrt2+2sqrt2+sqrt8)/sqrt3=5/3sqrt6#

Explanation:

As we have a surd in the denominator, here simplifying means rationalizing the denominator, which can be done by multiplying numerator and denominator by #sqrt3#.

Hence #(sqrt2+2sqrt2+sqrt8)/sqrt3#

= #(sqrt2+2sqrt2+sqrt(ul(2xx2)xx2))/sqrt3#

= #(sqrt2+2sqrt2+2sqrt2)/sqrt3#

= #(5sqrt2)/(sqrt3)xx(sqrt3)/(sqrt3)#

= #(5sqrt6)/3#

= #5/3sqrt6#

Aug 13, 2018

#sqrt8=sqrt4sqrt2=2sqrt2#

#[sqrt2+2sqrt2+2sqrt2]/sqrt3#

Collect like terms

#[5sqrt2]/(sqrt3)#

Rationalise the denominator

#[5sqrt2]/(sqrt3)xxsqrt3/sqrt3#

#[5sqrt6]/3#