How do you simplify #(sqrtx+sqrty)/(sqrtx-sqrty)#?

2 Answers
Narad T.
Feb 9, 2017

The answer is #=(x+y+2sqrt(xy))/(x-y)#

Explanation:

Multiply numerator and denominator by #sqrtx+sqrty#

So,

#(sqrtx +sqrty)/(sqrtx-sqrty)#

#=(sqrtx +sqrty)/(sqrtx-sqrty)*(sqrtx+sqrty)/(sqrtx+sqrty)#

#=(x+y+2sqrt(xy))/(x-y)#

George C.
Feb 9, 2017

#(sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y)) = (x+2sqrt(xy)+y)/(x-y)#

Explanation:

To "rationalise" the denominator, multiply by its radical conjugate #sqrt(x)+sqrt(y)# as follows:

#(sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y)) = ((sqrt(x)+sqrt(y))(sqrt(x)+sqrt(y)))/((sqrt(x)-sqrt(y))(sqrt(x)+sqrt(y))#

#color(white)((sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y))) = (x+2sqrt(x)sqrt(y)+y)/(x-y)#

#color(white)((sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y))) = (x+2sqrt(xy)+y)/(x-y)#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
2534 views around the world
You can reuse this answer
Creative Commons License