Question

How do you rationalize the denominator and simplify `(8-6sqrt5)/sqrt12`?

Answer 1

`[8-6 sqrt5]/sqrt12 = [2(4-3sqrt5)]/sqrt(4xx3)`

`rArr [2(4-3sqrt5)]/[2sqrt3]`

`rArr [4-3sqrt5]/[sqrt3]` multiply D & N BY`sqrt3`

`rArr [sqrt3(4-3sqrt5)]/[sqrt3 sqrt3]`

`rArr[sqrt3(4-3sqrt5)]/3`