SOLUTION: Rationalize the denominator of sqrtx+sqrty/sqrtx-sqrty assuming x>or equalt to 0 and y > or equal to 0. a. x+2 sqrt xy+y/x-y b. sqrtx-sqrty/sqrtx+sqrty c x^2+y^2/x^2-y^2 I t

Algebra ->  Radicals -> SOLUTION: Rationalize the denominator of sqrtx+sqrty/sqrtx-sqrty assuming x>or equalt to 0 and y > or equal to 0. a. x+2 sqrt xy+y/x-y b. sqrtx-sqrty/sqrtx+sqrty c x^2+y^2/x^2-y^2 I t      Log On


   

Question 977060: Rationalize the denominator of sqrtx+sqrty/sqrtx-sqrty assuming x>or equalt to 0 and y > or equal to 0.
a. x+2 sqrt xy+y/x-y
b. sqrtx-sqrty/sqrtx+sqrty
c x^2+y^2/x^2-y^2
I thought it was x+y/x-y but I was wrong. Thank you
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
According to the time-honored algebra conventions you were taught as "order of operations",
sqrt(x) + sqrt(y) / sqrt(x) - sqrt(y) =sqrt%28x%29%2Bsqrt%28y%29%2Fsqrt%28x%29-sqrt%28y%29
I bet what you meant is
( sqrt(x) + sqrt(y) ) / ( sqrt(x) - sqrt(y) ) =%28sqrt%28x%29%2Bsqrt%28y%29%29%2F%28sqrt%28x%29-sqrt%28y%29%29 ,
with parentheses to indicate that the numerator and denominator need to be calculated first, before dividing.
To rationalize, you multiply both parts (numerator and denominator) times the "conjugate" of the denominator.
(The conjugate is almost the same thing, but with the sign changed).
It works because %28a-b%29%28a%2Bb%29=a%5E2-b%5E2%29 ,
so with a=sqrt%28x%29 and b=sqrt%28y%29 , you have
,
and your denominator ends up with no square roots.
%28sqrt%28x%29%2Bsqrt%28y%29%29%2F%28sqrt%28x%29-sqrt%28y%29%29===%28sqrt%28x%29%2Bsqrt%28y%29%29%5E2%2F%28%28sqrt%28x%29%29%5E2-%28sqrt%28y%29%29%5E2%29==%28x%2B2sqrt%28xy%29%2By%29%2F%28x-y%29=( x + 2sqrt(xy) + y ) / (x-y)
That is what you tried to write for choice a.
What you wrote is
x + 2sqrt(xy) + y / x - y =x+%2B+2sqrt%28xy%29+%2By+%2F+x+-+y+