Question

If `(sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3)) = a+sqrt(15)b` for integers `a` and `b` then what are `a` and `b` ?

Answer 1

`a=4"; "b=1`

Explanation:

Assumption: The question should be:

`(sqrt5+sqrt3)/(sqrt5-sqrt3)=a+sqrt(15)color(white)(.) b`

`color(blue)("Consider the left hand side (LHS) only")`

`color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx1) )`

`color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx (sqrt5+sqrt3)/(sqrt5+sqrt3) )`

`((sqrt5+sqrt3)(sqrt5+sqrt3))/(5-3)`

`(5+2sqrt(3xx5)+3)/2`

`8/2+(2sqrt(15))/2`

`4+sqrt15`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`4+sqrt15=a+sqrt15color(white)(..)b`

Thus `a=4"; "b=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Additional note")`

If the RHS `->a+sqrt(15bcolor(white)(..))`

The `b` is still 1

Answer 2

`a=4" "` and `" "b=1`

Explanation:

To rational the denominator `sqrt(5)-sqrt(3)` we can multiply both numerator and denominator by the radical conjugate `sqrt(5)+sqrt(3)`...

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

`color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = ((sqrt(5))^2+2sqrt(5)sqrt(3)+(sqrt(3))^2)/((sqrt(5))^2-(sqrt(3))^2)`

`color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (5+2sqrt(15)+3)/(5-3)`

`color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (8+2sqrt(15))/2`

`color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = 4+sqrt(15)`

So `a=4` and `b=1`