Question
I. 2y2 – 19y + 35 = 0
II. 4x2 – 16x + 15 = 0
In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.
Solution
I. 2y2 – 19y + 35 = 0 2y2 – 5y – 14 y + 35 = 0 y (2 y – 5) – 7(2 y – 5) = 0 (2 y – 5)(y – 7) = 0 y = 7, 5/2 II. 4x2 – 16x + 15 = 0 4x2 – 6x – 10 x + 15 = 0 2 x(2 x – 3) – 5(2 x – 3) = 0 (2 x – 3)(2x – 5 )=0 x = 5/2, 3/2 Hence, x ≤y
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