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 = `{(-1)/2, 3}`
C = {x/2x2 – x – 3 = 0}
∴ 2x2 – x – 3 = 0
∴ 2x2 – 3x + 2x – 3 = 0
∴ x(2x – 3) + 1(2x – 3) = 0
∴ (2x – 3) + 1(2x – 3) = 0
∴ x(2x – 3) (x + 1) = 0
∴ 2x – 3 = 0 or x + 1 = 0
∴ x = `3/2` or x = – 1
∴ C = `{-1, 3/2}`
A ∪ B ∪ C = `{-5/3,3/2} ∪ {(-1)/2, 3} ∪ {-1, 3/2}`
= `{(-5)/3, -1, (-1)/2, 3/2, 3}`
