Question
(i) 2xΒ² β x β 3 = 0 (ii) 2yΒ² β 6y + 4 =
0 In each of these questions, two equations (i) and (ii) are given. You have to solve both the equations and give answers:Solution
(i) 2xΒ² β x β 3 = 0 => 2xΒ² + 2x - 3x β 3 = 0 => 2x (x + 1) - 3 (x + 1) = 0 => (x + 1) (2x - 3) = 0 => x = 3/2, β 1 (ii) 2yΒ² β 6y + 4 = 0 => 2yΒ² β 2y β 4y + 4 = 0 => 2y (y β 1) β 4 (y β 1) = 0 => (y β 1) (2y β 4) = 0 => y = 1, 2 No relation can be established between x and y.
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