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

3 Answers
Mar 14, 2018

Answer:

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

Answer:

#(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)#