SOLUTION: Find the product (xy) if x+y+{{{ sqrt( x+y ) }}} = 20 and x-y+{{{sqrt ( x-y ) }}}=12.

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Question 1104242: Find the product (xy) if x+y++sqrt%28+x%2By+%29+ = 20 and x-y+sqrt+%28+x-y+%29+=12.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
.
It is assumed in the given equations  that  sqrt%28x%2By%29 >= 0  and  sqrt%28x-y%29 >= 0.


1.  Let u = sqrt%28x+%2B+y%29.  Then

    u%5E2+%2B+u = 20,   or

    u%5E2+%2B+u+-+20 = 0.

    Factor left side

    (u+5)*(u-4) = 0.

    As  just noticed above, sqrt%28x%2By%29 assumed to be >= 0.
    Therefore, only positive root  u = 4  works.

    Thus,  sqrt%28x%2By%29 = 4.   Then  x + y = 4^2 = 16.



2.  Similarly, Let v = sqrt%28x+-+y%29.  Then

    v%5E2+%2B+v = 12,   or

    v%5E2+%2B+v+-+12 = 0.

    Factor left side

    (v+4)*(v-3) = 0.

    As  just noticed above, sqrt%28x-y%29 assumed to be >= 0.
    Therefore, only positive root  v = 3  works.

    Thus,  sqrt%28x-y%29 = 3.   Then  x - y = 3^2 = 9.

    

3.  Thus we have two equations

    x + y = 16,
    x - y =  9,


    which implies  x= 12.5,  y= 3.5.



Answer.  The solution is  x= 12.5,  y= 3.5.  The product xy = 43.75.

Solved.


Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Let u+=+sqrt%28x%2By%29

Then the first equation is
u%5E2%2Bu+=+20
u%5E2%2Bu-20+=+0
%28u%2B5%29%28u-4%29+=+0

Then u+=+sqrt%28x%2By%29+=+-5 or u+=+sqrt%28x%2By%29+=+4.

Since the square root has to be positive, sqrt%28x%2By%29=4, which means x%2By=16 (1).

Similarly, let u+=+sqrt%28x-y%29; then the second equation is

u%5E2%2Bu+=+12
u%5E2%2Bu-12+=+0
%28u%2B4%29%28u-3%29+=+0

Then u+=+sqrt%28x-y%29+=+-4 or u+=+sqrt%28x-y%29+=+3.

Then, as before, sqrt%28x-y%29+=+3, which means x-y+=+9 (2).

Solving (1) and (2) together gives us x = 25/2, y = 7/2.

Then the product we are looking for is %2825%2F2%29%2A%287%2F2%29+=+175%2F4