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Question
If A = {x/6x2 + x - 15 = 0}
B = {x/2x2 - 5x - 3 = 0}
C = {x/2x2 - x - 3 = 0} then
find (A ∩ B ∩ C)
Solution
A = {x/6x2 + x - 15 = 0}
∴ 6x2 + x - 15 = 0
∴ 6x2 + 10x - 9x - 15 = 0
∴ 2x(3x + 5) - 3(3x + 5) = 0
∴ (3x + 5)(2x - 3) = 0
∴ 3x + 5 = 0 or 2x - 3 = 0
∴ x = `(-5)/3` or x = `3/2`
∴ A = `{(-5)/3,3/2}`
B = {x/2x2 - 5x - 3 = 0}
∴ 2x2 - 5x - 3 = 0
∴ 2x2 - 6x + x - 3 = 0
∴ 2x(x - 3) + 1(x - 3) = 0
∴ (x - 3)(2x + 1) = 0
∴ x - 3 = 0 or 2x + 1 = 0
∴ x = 3 or x = `(-1)/2`
∴ B = `{3,(-1)/2}`
C = {x/2x2 - x - 3 = 0}
∴ 2x2 - x - 3 = 0
∴ 2x2 - 3x + 2x - 3 = 0
∴ x(2x - 3) + 1(2x - 3) = 0
∴ (2x - 3)(x + 1) = 0
∴ 2x - 3 = 0 or x + 1 = 0
∴ x = `3/2` or x = -1
∴ c = `{3/2, -1}`
A ∩ B ∩ C = `{(-5)/3,3/2} ∩ {3,(-1)/2} ∩ {3/2, -1}`
A ∩ B ∩ C = ϕ
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