How do you simplify #(12+sqrt108)/6#?

2 Answers
May 29, 2017

Answer:

#= 2+sqrt3#

Explanation:

Each term in the numerator must be divided by #6:#

#(12+sqrt108)/6 = 12/6+sqrt108/6" "larr# prime factor 108

#=2+ sqrt(2xx2xx3 xx3xx3)/6 = 2+sqrt(2^2xx3^2xx3)/6#

Find the roots where possible.

#2+(2*3sqrt3)/6 = 2+(cancel6sqrt3)/cancel6#

#= 2+sqrt3#

May 29, 2017

Answer:

#2+sqrt3#

Explanation:

#"simplifying " sqrt108" to begin with"#

#"using the "color(blue)"law of radicals"#

#color(orange)"Reminder " sqrtaxxsqrtbhArrsqrt(ab)#

#rArrsqrt108=sqrt(2^2xx3^3)=2xx3sqrt3=6sqrt3#

#rArr(12+sqrt108)/6#

#=(12+6sqrt3)/6#

#=12/6+(6sqrt3)/6#

#=2+sqrt3#