SOLUTION: x^2+y^2=48 (x-y)=4 Find the value of xy

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Question 658834: x^2+y^2=48
(x-y)=4
Find the value of xy
Found 2 solutions by Alan3354, kevwill:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=48
(x-y)=4
Find the value of xy
---------
x = y+4
(y+4)^2 + y^2 = 48
etc

Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
From x+-+y+=+4 we know y+=+x-4
Substituting into the first equation gives
x%5E2+%2B+%28x-4%29%5E2+=+48
Expanding:
x%5E2+%2B+x%5E2+-+8x+%2B+16+=+48
Combining like terms:
2x%5E2+-+8x+%2B+16+=+48
Subtracting 48 from both sides:
2x%5E2+-+8x+-+32+=+0
Dividing both sides by 2:
x%5E2+-+4x+-+16+=+0
This can be solved using the quadratic equation x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ with a=1, b=-4, and c=-16
x+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A1%2A%28-16%29+%29%29%2F%282%2A1%29+
x+=+%284+%2B-+sqrt%28+16%2B64+%29%29%2F2+
x+=+%284+%2B-+sqrt%28+80+%29%29%2F2+
x+=+%284+%2B-+sqrt%28+16%2A5+%29%29%2F2+
x+=+%284+%2B-+4%2Asqrt%28+5+%29%29%2F2+
x+=+2+%2B-+2%2Asqrt%28+5+%29+
So x can have the values x=2%2B2%2Asqrt%285%29 or x=2-2%2Asqrt%285%29
For x+=+2%2B2%2Asqrt%285%29 we have y+=+x-4+=+-2%2B2%2Asqrt%285%29-2

Checking:

And

For x+=+2-2%2Asqrt%285%29 we have y+=+x-4+=+-2-2%2Asqrt%285%29
Checking:

And

So for either value of x, x*y = 16