Question
I. 3y2 + 16y + 16 = 0 II.
2x2 + 19x + 45 = 0Solution
I. 3y2 + 16y + 16 = 0 3y2 + 4y + 12y + 16 = 0 y (3 y + 4) + 4(3 y + 4) = 0 (y + 4)(3 y + 4) = 0 y = - 4, -4/3 II. 2x2 + 19x + 45 =0 2x2 + 9x + 10 x + 45 =0 x (2 x + 9) + 5 (2 x + 9) = 0 (x + 5) (2 x + 9) = 0 x = - 5, -9/2 Hence, x > y
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