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

3 Answers
Mar 14, 2018

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)

Mar 14, 2018

(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 :)

Mar 14, 2018

(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)