How do you simplify this? #" "sqrt(3) + sqrt(1/3)#

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Answer 1 · 2018-05-21T14:58:59

#sqrt3+sqrt(1/3)=4/sqrt3#

#sqrt3+sqrt(1/3)#

#sqrt3=3/sqrt3=#

#sqrt(1/3)=1/sqrt3#

#sqrt3+sqrt(1/3)=3/sqrt3+1/sqrt3#
#=(3+1)/sqrt3=4/sqrt3#
Thus,

#sqrt3+sqrt(1/3)=4/sqrt3#

Answer 2 · 2018-05-21T15:00:40

#=4/3 sqrt3#

You can write the square root of the fraction as separate roots:

#sqrt3 + color(blue)(sqrt(1/3)) = sqrt3 + color(blue)(sqrt1/sqrt3)#

#sqrt3 + 1/sqrt3" "larr# rationalise the denominator

#=sqrt3 +1/sqrt3 xx sqrt3/sqrt3#

#=sqrt3 +sqrt3/3" "larr# factor out #sqrt3#

#=sqrt3(1+1/3)#

#=4/3 sqrt3#

Answer 3 · 2018-05-21T16:14:39

#sqrt3+ sqrt(1/3)# is not an equation and, therefore, cannot be solved but it can be rationalized.

#sqrt3+ sqrt(1/3)#

Separate the second term into a radical over a radical:

#sqrt3+ sqrt1/sqrt3#

Multiply the second term by 1 in the form of #sqrt3/sqrt3#

#sqrt3+ sqrt1/sqrt3sqrt3/sqrt3#

Because of #sqrt3sqrt3=3#: the denominator becomes 3:

#sqrt3+ sqrt3/3#

Multiply the first term by 1 in the form of #3/3#:

#(3sqrt3)/3+ sqrt3/3#

The two fractions can be combined over the common denominator:

#(3sqrt3+ sqrt3)/3#

Add the terms in the numerator

#(4sqrt3)/3#

The above is the rationalized and simplified form.