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Math Help - is there a trick to solving inequalities?

  1. #1
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    is there a trick to solving inequalities?

    I have some problems of which i solved them all by pretty much factoring and then guess and check.
    x(x-1)>0
    4x^2+10x-6<0
    x^2 +2x+4>0 (I couldnt solve this one because I couldnt factor it)

    So if you guys know a trick to solving these please tell me because guess/check is making my brain fry. Also If you know how to factor x^2+2x+4>0 please tell me that would rule. (or atleast how to solve it as an inequality if you cant factor it)
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  2. #2
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    Quote Originally Posted by frankinaround View Post
    I have some problems of which i solved them all by pretty much factoring and then guess and check.
    x(x-1)>0
    4x^2+10x-6<0
    x^2 +2x+4>0 (I couldnt solve this one because I couldnt factor it)

    So if you guys know a trick to solving these please tell me because guess/check is making my brain fry. Also If you know how to factor x^2+2x+4>0 please tell me that would rule. (or atleast how to solve it as an inequality if you cant factor it)
    think about the graph of the quadratic (a parabola) which opens upward for each of your examples ...

    y = x(x-1) crosses the x-axis at x = 0 and x = 1 ... y > 0 when x < 0 and when x > 1

    y = 4x^2 + 10x + 6 = 2(2x^2 + 5x + 3) = 2(2x + 3)(x + 1) crosses the x-axis at x = -3/2 and x = -1 ... y < 0 for -3/2 < x < -1

    x^2 + 2x + 4 = (x^2 + 2x + 1) + 3 = (x + 1)^2 + 3 > 0 for all x
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  3. #3
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    Quadratic inequalities are easiest solved if you complete the square.

    1. \displaystyle x(x-1) > 0

    \displaystyle x^2 - x > 0

    \displaystyle x^2 - x + \left(-\frac{1}{2}\right)^2 > \left(-\frac{1}{2}\right)^2

    \displaystyle \left(x - \frac{1}{2}\right)^2 > \frac{1}{4}

    \displaystyle \left|x - \frac{1}{2}\right| > \frac{1}{2}

    \displaystyle x - \frac{1}{2} < -\frac{1}{2} or \displaystyle x - \frac{1}{2} > \frac{1}{2}

    \displaystyle x < 0 or \displaystyle x > 1.


    2. \displaystyle 4x^2 + 10x - 6 < 0

    \displaystyle x^2 + \frac{5}{2}x - \frac{3}{2} < 0

    \displaystyle x^2 + \frac{5}{2}x + \left(\frac{5}{4}\right)^2 - \frac{3}{2} < \left(\frac{5}{4}\right)^2

    \displaystyle \left(x + \frac{5}{4}\right)^2 - \frac{24}{16} < \frac{25}{16}

    \displaystyle \left(x + \frac{5}{4}\right)^2 < \frac{49}{16}

    \displaystyle \left|x + \frac{5}{4}\right|^2 < \frac{7}{4}

    \displaystyle -\frac{7}{4} < x + \frac{5}{4} < \frac{7}{4}

    \displaystyle -3 < x < \frac{1}{2}.


    3. \displaystyle x^2 + 2x + 4 > 0

    \displaystyle x^2 + 2x + 1^2 + 4 > 1^2

    \displaystyle (x + 1)^2 + 4 > 1

    \displaystyle (x + 1)^2 > -3


    Since the LHS is always nonnegative, this inequality is true for all \displaystyle x.
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