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If A = {x/6x2 + x – 15 = 0}, B = {x/2x2 – 5x – 3 = 0}, C = {x/2x2 – x – 3 = 0} then find (A ∪ B ∪ C). - Mathematics and Statistics

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प्रश्न

If A = {x/6x2 + x – 15 = 0}, B = {x/2x2 – 5x – 3 = 0}, C = {x/2x2 – x – 3 = 0} then find (A ∪ B ∪ C).

योग

उत्तर

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}`

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अध्याय 5: Sets and Relations - Exercise 5.1 [पृष्ठ ९७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 5 Sets and Relations
Exercise 5.1 | Q 3. (i) | पृष्ठ ९७

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