The number of ordered pairs (x, y) satisfying the equation x^2 + 2x (xy) + 1 = 0 , where y [0, 3 ] , is:

Step-by-step solution

Given equation: x^2 + 2x (xy) + 1 = 0 This can be rewritten by completing the square: (x^2 + 2x (xy) + ^2(xy)) + 1 - ^2(xy) = 0 (x + (xy))^2 + ^2(xy) = 0 Since the sum of two non-negative quantities is zero, each must be zero individually: (x + (xy))^2 = 0 x = - (xy) ^2(xy) = 0 (xy) = 0 From (xy) = 0 , we have (xy) = 1 or (xy) = -1 . Case 1: (xy) = 1 Then x = -1 . Substituting x = -1 into (xy) = 1 , we get (-y) = 1 (y) = 1 . For y [0, 3 ] , the solutions are y = 0, 2 . This gives the ordered pairs (-1, 0) and (-1,