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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} is the set of the solutions i.e., the solution set of the equation 6x2 + x – 15 = 0. Hence we solve this equation.

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`

∴ the solution set of the equation

∴ 6x2 + x – 15 = 0 is A = `{-5/3, 3/2}`.

Similarly, solving the equations 2x2 – 5x – 3 = 0 and 2x2 – x – 3 = 0,

we have B = `{-1/2, 3}` and C = `{3/2, -1}`.

Hence,
Since the sets A, B and C have no common element, A ∩ B ∩ C = `phi`

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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. (ii) | पृष्ठ ९७

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