One Solution
(x^2-4x+7)*(y^2+2y+6) = 15
(x^2-4x+4 + 3)*(y^2+2y+1 + 5) = 15
((x-2)^2 + 3)*((y+1)^2 + 5) = 15
if (x,y) = (2,-1) then:
3 * 5 = 15
So one solution is (x,y) = (2,-1) and requested product x*y = -2
One could set the pair of quadratics to all the combination of factors that multiply to 15:
(x^2-4x+7) , (y^2+2y+6) = 3 , 5 has already been done above
But there is still:
A , B
1 , 15
5 , 3
15 , 1
-1 , -15
-3 , -5
-5 , -3
-15 , -1
Each would produce 2 roots for f(x) = A and 2 roots for g(y) = B, potentially 4 values for x*y.
Will do for the first example of 1,15
x^2-4x+7 = 1 y^2+2y+6 = 15
x^2-4x+6 = 0 y^2+2y-9 = 0
x roots: 2 ± √2 i y roots: -1 ± √10
There are 4 different complex values for an x root times a y root; but the arithmetic average is still -2.
I suspect this holds true for the other examples but I have not checked.
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Posted by Larry
on 2024-05-08 13:02:30 |
One Solution
