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Find all ordered pairs of real numbers (x,y) such that x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0 .

waffles  Mar 24, 2018
 #1
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This can be rearranged as (x+1)^2 = -2y^2/(y^2+5)

 

The right hand side is negative unless y is zero, whereas the left hand side is positive unless x = -1.

 

The only way to achieve equality is for both sides to be zero, which means the only ordered pair of real numbers to satisfy the equation is (x,y) = (-1,0)

Alan  Mar 25, 2018

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