Question

How do you rationalize the denominator and simplify `5/(sqrt3-1)`?

Answer 1

To rationalise this denominator, you have to take the denominator's conjugate. The conjugate will evaluate into a difference of two squares ( `a^2 - b^2 = (a + b)(a - b)` ) which can be expressed as an integer.

So to rationalise this expression, `5/(sqrt(3)-1)`

`5/(sqrt(3)-1)`

= `5/(sqrt(3)-1) * (sqrt(3)+1)/(sqrt(3)+1)`

= `(5(sqrt(3)+1))/(3-1)`

= `(5sqrt(3) + 5)/(2)`

Answer 2

`color(green)((5(sqrt3+1))/2`

Explanation:

`:.5/(sqrt3-1) xx (sqrt3+1)/ (sqrt3+1)`

`sqrt3 xx sqrt3=3`

`color(white)(aaaaaaaaaaaaa)` `sqrt3-1`
`color(white)(aaaaaaaaaaa)` `xx underline(sqrt3+1)`
`color(white)(aaaaaaaaaaaaa)` `3-sqrt3`
`color(white)(aaaaaaaaaaaaaaaa)` `sqrt3-1`
`color(white)(aaaaaaaaaaaaa)` `overline(3+0-1)`

`color(white)(aaaaaaaaaaaaa)` `color(green)(=2`

`:.color(green)((5(sqrt3+1))/2`

~~~~~~~~~~~~~~~~~~~~~

check by calculator:

`:.5/(sqrt3-1)=color(green)(6.830127019`

`:.(5(sqrt3+1))/2=color(green)(6.830127019`