Question

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

Answer 1

`= 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`

Answer 2

`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`