Question

How do you rationalize `(2sqrt5-8)/ (2sqrt5+3)`?

Answer 1

`2(2-sqrt5)`

Explanation:

`(2 sqrt5-8)/(2sqrt5+3)`. Multiplying by `(2sqrt5-3)` on

both numerator and denominator we get,

`=((2 sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))`

`=(20-2sqrt5(8+3)+24)/((2sqrt5)^2-3^2)`

`=(44-22sqrt5)/(20-9)=(22(2-sqrt5))/11`

`=2(2-sqrt5)` [Ans]

Answer 2

`(2sqrt5-8)/(2sqrt5+3)=4-2sqrt5`

Explanation:

To rationalize the denominator, we multiply by the conjugate and use the difference of squares rule. In this case, the conjugate is `2sqrt5-3`, so we multiply by it on both top and bottom:

`(2sqrt5-8)/(2sqrt5+3)=((2sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))`

The difference of squares rule says:
`(a+b)(a-b)=a^2-b^2`

Applying this to the denominator, we get:
`((2sqrt5-8)(2sqrt5-3))/(4*5-3)`

Then we multiply out the top:
`(20-6sqrt5-16sqrt5+24)/11=(44-22sqrt5)/11=4-2sqrt5`