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Math Help - Quadratic Inequality

  1. #1
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    Quadratic Inequality

    How would I solve the following quadratic inequalities?

    1. \frac{x^{2} - 2x + 3}{x^{2} - 2x} < 0

    2. \frac{3x + 4}{x^{2} -3x + 2} \geq  0
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  2. #2
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    Hello SVXX
    Quote Originally Posted by SVXX View Post
    How would I solve the following quadratic inequalities?

    1. \frac{x^{2} - 2x + 3}{x^{2} - 2x} < 0

    2. \frac{3x + 4}{x^{2} -3x + 2} \geq  0
    1. Note that x^2 -2x +3 = (x-1)^2 +2, which is always positive. So you just need to consider the sign of the denominator, x(x - 2), in the each of the ranges x<0, 0<x<2, x>2.

    2. \frac{3x + 4}{x^{2} -3x + 2} = \frac{3x+4}{(x-1)(x-2)}\ge 0

    Consider the values of x which make each term zero; these are -\tfrac43, 1, 2, and look at the sign of the expression as x moves along the number line from left to right through these values; in other words in the ranges

    • x<-\tfrac43
    • -\tfrac43 < x < 1
    • 1<x<2
    • x>2

    For instance, when x<-\tfrac43, you get \frac{-^{ve}}{-^{ve}\times -^{ve}}=-^{ve}. So x<-\tfrac43 is not part of the solution set.

    Grandad
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  3. #3
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    Aha..now I finally understand.
    Also, in the second part, the checkpoint -4/3 will be part of the solution as it makes the expression = 0 but 1 and 2 will not be part of the solution as they make the expression indeterminate.
    Thanks a bunch
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