How to solve this quadratic...please help
- Thread starter smelius
- Start date
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Do you mean (12x^2 + 11x + 2) / (x - 6) > 0? Parentheses are important there.
(12x<sup>2</sup> + 11x + 2) / (x - 6) > 0
<==> (12x<sup>2</sup> + 11x + 2 > 0 and x - 6 > 0) or (12x<sup>2</sup> + 11x + 2 < 0 and x - 6 < 0)
<==> ((3x + 2)(4x + 1) > 0 and x - 6 > 0) or ((3x + 2)(4x + 1) < 0 and x - 6 < 0)
<==> (((3x + 2 > 0 and 4x + 1 > 0) or (3x + 2 < 0 and 4x + 1 < 0)) and x - 6 > 0) or (((3x + 2 > 0 and 4x + 1 < 0) or (3x + 2 < 0 and 4x + 1 > 0)) and x - 6 < 0)
<==> (((3x > -2 and 4x > -1) or (3x < -2 and 4x < -1)) and x > 6) or (((3x > -2 and 4x < -1) or (3x < -2 and 4x > -1)) and x < 6)
<==> (((x > -2/3 and x > -1/4) or (x < -2/3 and x < -1/4)) and x > 6) or (((x > -2/3 and x < -1/4) or (x < -2/3 and x > -1/4)) and x < 6)
<==> ((x > -1/4 or x < -2/3) and x > 6) or (x > -2/3 and x < -1/4 and x < 6)
<==> x > 6 or -2/3 < x < -1/4
(12x<sup>2</sup> + 11x + 2) / (x - 6) > 0
<==> (12x<sup>2</sup> + 11x + 2 > 0 and x - 6 > 0) or (12x<sup>2</sup> + 11x + 2 < 0 and x - 6 < 0)
<==> ((3x + 2)(4x + 1) > 0 and x - 6 > 0) or ((3x + 2)(4x + 1) < 0 and x - 6 < 0)
<==> (((3x + 2 > 0 and 4x + 1 > 0) or (3x + 2 < 0 and 4x + 1 < 0)) and x - 6 > 0) or (((3x + 2 > 0 and 4x + 1 < 0) or (3x + 2 < 0 and 4x + 1 > 0)) and x - 6 < 0)
<==> (((3x > -2 and 4x > -1) or (3x < -2 and 4x < -1)) and x > 6) or (((3x > -2 and 4x < -1) or (3x < -2 and 4x > -1)) and x < 6)
<==> (((x > -2/3 and x > -1/4) or (x < -2/3 and x < -1/4)) and x > 6) or (((x > -2/3 and x < -1/4) or (x < -2/3 and x > -1/4)) and x < 6)
<==> ((x > -1/4 or x < -2/3) and x > 6) or (x > -2/3 and x < -1/4 and x < 6)
<==> x > 6 or -2/3 < x < -1/4