Question
I. 2y² - 11 y + 15 = 0 II. 2x² + 3x
– 14 = 0Solution
I. 2y² - 11 y + 15 = 0 2y² - 5y - 6 y + 15 = 0 y (2 y - 5) – 3(2 y - 5) = 0 (y – 3) (2 y - 5) = 0 y = 3, 5/2 II. 2x² + 3x – 14 = 0 2x² + 7x - 4 x – 14 = 0 x (2 x + 7) – 2 (2 x + 7) = 0 (x – 2) (2 x + 7) = 0 x = 2, - 7/2 Hence, x<y
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