Question

How do you write this easier ?

`(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2)`

Answer 1

See a solution process below:

Explanation:

We can rationalize the denominator by multiplying this expression by the appropriate form of `1`

We can use this rule of quadratics to find the appropriate form of `1` to eliminate the radicals from the denominator:

`(color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - color(blue)(y)^2`

And we can use this rule of quadratics to multiply the numerator:

`(color(red)(x) - color(blue)(y))(color(red)(x) - color(blue)(y)) = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2`

`(color(red)(sqrt(3)) - color(blue)(sqrt(2)))/(color(red)(sqrt(3)) - color(blue)(sqrt(2))) xx (color(red)(sqrt(3)) - color(blue)(sqrt(2)))/(color(red)(sqrt(3)) + color(blue)(sqrt(2))) =>`

`(color(red)(sqrt(3))^2 - 2color(red)(sqrt(3))color(blue)(sqrt(2)) + color(blue)(sqrt(2))^2)/(color(red)(sqrt(3))^2 - color(blue)(sqrt(2))^2) =>`

`(3 - 2sqrt(3 * 2) + 2)/(3 - 2) =>`

`(3 + 2 - 2sqrt(6))/1 =>`

`5 - 2sqrt(6)`

Answer 2

`5-2sqrt6`

Explanation:

`"what we have to do here is "color(blue)"rationalise the denominator"`

`"this means eliminating the radicals on the denominator"`
`"and replacing them with a rational value "`

`"this is achieved by multiplying the numerator and"`
`"denominator by the "color(blue)"conjugate ""of the denominator"`

`"the conjugate of "sqrt3+sqrt2" is "sqrt3color(red)(-)sqrt2`

`"note " sqrtaxxsqrta=a" and "sqrtaxxsqrtbhArrsqrt(ab)`

`rArr(sqrt3+sqrt2)(sqrt3-sqrt2)larr" expand using FOIL"`

`=3cancel(-2sqrt3)cancel(+2sqrt3)-2`

`=1larrcolor(red)" rational value"`

`rArr(sqrt3-sqrt2)/(sqrt3+sqrt2)`

`=((sqrt3-sqrt2)(sqrt3-sqrt2))/((sqrt3+sqrt2)(sqrt3-sqrt2))`

`=(3-sqrt6-sqrt6+2)/1`

`=5-2sqrt6`