Question
I. 2x2 – 25x + 33 = 0 II.
3y2 + 40y + 48 = 0Solution
I. 2x2 – 25x + 33 = 0 2x2 – 3 x – 22x + 33 = 0 x (2 x – 3) – 11(2 x – 3) = 0 (x – 11) (2 x – 3) = 0 x = 11, 3/2 II. 3y2 + 40y + 48 = 0 3y2 + 36y + 4 y + 48 = 0 3 y(y + 12) + 4(y + 12) = 0 (y + 12) (3 y + 4) = 0 y = – 12, -4/3 Hence, x > y
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