How do you rationalize the denominator and simplify #(sqrtx + 2sqrty)/(sqrtx - 2sqrty)#?

Question

1547 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-03T18:31:49

#(x+4sqrt(xy)+4y)/(x-4y)#

Multiply the fraction by the conjugate of its denominator.

#=(sqrtx+2sqrty)/(sqrtx-2sqrty)((sqrtx+2sqrty)/(sqrtx+2sqrty))#

In the denominator, we have what will become a difference of squares, which take the form:

#(a+b)(a-b)=a^2-b^2#

Here, we have #a=sqrtx#, so #a^2=x#, and #b=2sqrty#, which implies that #b^2=4y#.

Thus, the fraction simplifies to be

#=(sqrtx+2sqrty)^2/(x-4y)#

This is a simplified answer. However, another perfectly acceptable way of presenting this would be to distribute the squared binomial as follows:

#=((sqrtx+2sqrty)(sqrtx+2sqrty))/(x-4y)=(x+4sqrt(xy)+4y)/(x-4y)#