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Thread: Quadratic Inequality

  1. #1
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    Quadratic Inequality

    How would I solve the following quadratic inequalities?

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

    2.$\displaystyle \frac{3x + 4}{x^{2} -3x + 2} \geq 0$
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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.$\displaystyle \frac{x^{2} - 2x + 3}{x^{2} - 2x} < 0$

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

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

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

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

    For instance, when $\displaystyle x<-\tfrac43$, you get $\displaystyle \frac{-^{ve}}{-^{ve}\times -^{ve}}=-^{ve}$. So $\displaystyle x<-\tfrac43$ 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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