Question
I. 3x² - 22 x + 40 = 0  II. 4y² + 22y +
24 = 0Â Â Â In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.Solution
I. 3x² - 22 x + 40 = 0   3x² - 12x - 10x + 40 = 0   3x (x – 4) – 10 (x – 4) (3x – 10 ) (x – 4) x = 10/3, 4 II. 4y² + 22y + 24 = 0  4y² + 16 y + 6 y + 24 = 0   4y (y + 4) + 6( y + 4) (4y + 6) (y + 4) x = - 4 , - 3/2 Hence, x > y.
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