I. x2 + 16x + 63 = 0
II. y2 + 2y - 15 = 0
I. x2 + 16x + 63 = 0 => x2 + 9x + 7x + 63 = 0 => x(x + 9) + 7(x + 9) = 0 => (x + 9) (x +7) = 0 => x = -9, -7 II. y2 + 2y - 15 = 0 => y2 + 5y – 3y - 15 = 0 => y(y + 5) – 3(y + 5) = 0 => (y + 5) (y - 3) = 0 => y = -5, 3 Hence, x < y.
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