Question

How do you simplify `(sqrta- sqrtb)/(sqrta+sqrtb)`?

Answer 1

See a solution process below:

Explanation:

To simplify we need to rationalize the denominator by multiplying by the appropriate form of `1`:

`(color(red)(sqrt(a)) - color(red)(sqrt(b)))/(color(red)(sqrt(a)) - color(red)(sqrt(b))) xx (sqrt(a) - sqrt(b))/(sqrt(a) + sqrt(b)) =>`

`(sqrt(a) - sqrt(b))^2/(color(red)(sqrt(a))sqrt(a) + color(red)(sqrt(a))sqrt(b) - color(red)(sqrt(b))sqrt(a) - color(red)(sqrt(b))sqrt(b)) =>`

`(sqrt(a) - sqrt(b))^2/(a - b)`

Answer 2

`(sqrta-sqrtb)/(sqrta+sqrtb)`

Multiply both the numerator and denominator by `sqrta-sqrtb`

`(sqrta-sqrtb)/(sqrta+sqrtb)*(sqrta-sqrtb)/(sqrta-sqrtb)`

`(sqrta-sqrtb)^2/((sqrta)^2-(sqrtb)^2)`

`(a+b-2sqrt(ab))/(a-b)`

Hope this helps :)

Answer 3

`(a-2sqrt(ab)+b)/(a-b)`

Explanation:

`"multiply the numerator/denominator by the "color(blue)"conjugate"`
`"of the denominator"`

`"the conjugate of "sqrta+sqrtb" is "sqrtacolor(red)(-)sqrtb`

`"this ensures the denominator is a rational value"`

`•color(white)(x)sqrtaxxsqrta=a`

`•color(white)(x)(sqrta+sqrtb)(sqrta-sqrtb)=a-b`

`rArr(sqrta-sqrtb)/(sqrta+sqrtb)xx(sqrta-sqrtb)/(sqrta-sqrtb)`

`=((sqrta-sqrtb)(sqrta-sqrtb))/((sqrta+sqrtb)(sqrta-sqrtb))`

`=(a-sqrtab-sqrtab+b)/(a-b)`

`=(a-2sqrt(ab)+b)/(a-b)`