Question

How do you rationalize the denominator and simplify `(x-3)/(sqrtx-sqrt3)`?

Answer 1

To rationalize a denominator in the form of `sqrta - sqrtb`, you multiply the fraction by 1 in the form `(sqrta + sqrtb)/(sqrta + sqrtb)`

Explanation:

The reason for doing this practice comes from general form for factoring binomials that contain the difference two squares:

`a^2 - b^2 = (a - b)(a + b)`

Returning to the given fraction, we multiply by 1 in form `(sqrtx + sqrt3)/(sqrtx + sqrt3)`

`(x - 3)/(sqrtx - sqrt3)(sqrtx + sqrt3)/(sqrtx + sqrt3) =`

`((x - 3)(sqrtx + sqrt3))/(x - 3) =`

`sqrtx + sqrt3`

Answer 2

`sqrt x + sqrt 3`

Explanation:

divide the Numerator and denominator by `sqrtx + sqrt 3`.
we get, `( x - 3)/(sqrt x - sqrt 3) * (sqrt x + sqrt 3)/(sqrt x + sqrt 3)`
= #[(x - 3)(sqrt x + sqrt 3)]/[(sqrt x)^2 - (sqrt 3)^2] = [(x - 3)(sqrt x + sqrt 3)]/(x - 3) = sqrt x + sqrt 3#