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Roots and products (Posted on 2024-05-08) Difficulty: 3 of 5
Find ordered pairs (x, y) that solve the equation

(x2-4x+7)(y2+2y+6)=15

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts One Solution | Comment 1 of 2
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.

  Posted by Larry on 2024-05-08 13:02:30
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