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Math Help - How do I solve this inequality?

  1. #1
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    How do I solve this inequality?

    Hi

    I've gone through the "Solving inequalities" article. However, I didn't find anything in the document or my textbooks that could help me solve this inequality:

    [(|x|-3)/ (x-2)] < 2x

    Could someone please guide me in solving this?

    Thanks in advance.
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  2. #2
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    \mbox{multiply (x-2) and then add 3} \ \ |x|<2x(x-2)+3\Rightarrow |x|<2x^2-4x+3

    \Rightarrow x<2x^2-4x+3 \ \mbox{and} \ -x<2x^2-4x+3\Rightarrow -2x^2+4x-3<x<2x^2-4x+3

    |x| \ \mbox{is} \ -x \ \forall x\in\mathbb{R}^- \ \mbox{and} \ x \ \forall x\in\mathbb{R}^+
    Last edited by dwsmith; December 23rd 2010 at 10:37 AM.
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  3. #3
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    You need to consider a few cases, since when you multiply by a negative number, the inequality sign changes direction.

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

    First, it should be clear that \displaystyle x - 2 \neq 0 \implies x \neq 2.


    Case 1: \displaystyle x - 2 > 0 \implies x > 2, then

    \displaystyle |x| - 3 < 2x(x-2)

    \displaystyle |x| - 3 < 2x^2 - 4x.


    Since \displaystyle x > 2 and is therefore positive, we can say \displaystyle |x| = x, so

    \displaystyle x - 3 < 2x^2 - 4x

    \displaystyle 2x^2 - 4x > x - 3

    \displaystyle 2x^2 - 5x + 3 > 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 - \left(-\frac{5}{4}\right)^2 + \frac{3}{2} > 0

    \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{25}{16} + \frac{24}{16} > 0

    \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{1}{16} > 0

    \displaystyle \left(x - \frac{5}{4}\right)^2 > \frac{1}{16}

    \displaystyle \left|x - \frac{5}{4}\right| > \frac{1}{4}

    \displaystyle x - \frac{5}{4} < -\frac{1}{4} or \displaystyle x - \frac{5}{4} > \frac{1}{4}

    \displaystyle x < 1 or \displaystyle x > \frac{3}{2}.


    Since we know that \displaystyle x > 2, we can say \displaystyle x > 2 satisfies the original inequality.


    Case 2: \displaystyle x - 2 < 0 \implies x < 2, then

    \displaystyle |x| - 3 > 2x(x - 2)

    \displaystyle |x| - 3 > 2x^2 - 4x.


    Since \displaystyle x < 2, we need to consider that \displaystyle |x| = x if \displaystyle 0 \leq x < 2 and \displaystyle |x| = -x if \displaystyle x < 0.


    Case 2a) \displaystyle 0 \leq x < 2 \implies |x| = x, then

    \displaystyle x - 3 > 2x^2 - 4x

    \displaystyle 2x^2 - 4x < x - 3

    \displaystyle 2x^2 - 5x + 3 < 0

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

    \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{1}{16} < 0

    \displaystyle \left(x - \frac{5}{4}\right)^2 < \frac{1}{16}

    \displaystyle \left|x - \frac{5}{4}\right| < \frac{1}{4}

    \displaystyle -\frac{1}{4} < x - \frac{5}{4} < \frac{1}{4}

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


    Since we know \displaystyle 0 \leq x <2, that means \displaystyle 1 < x < \frac{3}{2} satisfies the inequality.


    Case 2b) \displaystyle x < 0 \implies |x| = -x, then

    \displaystyle -x - 3 > 2x^2 - 4x

    \displaystyle 2x^2 - 4x < -x - 3

    \displaystyle 2x^2 - 3x + 3 < 0

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

    \displaystyle x^2 - \frac{3}{2}x + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 + \frac{3}{2} < 0

    \displaystyle \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + \frac{24}{16} < 0

    \displaystyle \left(x - \frac{3}{4}\right)^2 + \frac{15}{16} < 0.

    Since the LHS is always positive, this inequality is never true.



    So putting it all together, the solution to the inequality

    \displaystyle \frac{|x| - 3}{x - 2} < 2x is

    \displaystyle x\in \left(1, \frac{3}{2}\right) \cup \left(2, \infty\right).
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  4. #4
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    Thanks for the help.

    I'm sorry that I stated that I was looking for the solution to this inequality. What we were supposed to find out was the range of x.

    However the answer to this is (3-33^0.5)/4 < x < 2 or x < 3.

    There's one more thing that I find different from your approaches. In the text, it's stated that to remove the denominator from either the LHS or RHS, we have to multiply the square of the denominator to both sides. However, the steps shown above just require me to bring the denominator over to the other side.

    Are both these methods one and the same?
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  5. #5
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    dwsmith,
    can you multiply both sides by (x-2) without knowing whether it is positive or not?
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  6. #6
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    Quote Originally Posted by BAdhi View Post
    dwsmith,
    can you multiply both sides by (x-2) without knowing whether it is positive or not?
    Could this be the reason why we have to multiply the square of the denominator on both sides to remove it?
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  7. #7
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    Alternatively...

    \displaystyle\frac{|x|}{x-2}-\frac{3}{x-2}<2x\Rightarrow\ \frac{|x|}{x-2}<\frac{3}{x-2}+2x\Rightarrow\frac{|x|}{x-2}<\frac{2x^2-4x+3}{x-2},\;\;\;x\ \ne\ 2

    The 3 cases are

    (1)\;\;x>2,\;\;\;(2)\;\;0<x<2,\;\;\;(3)\;\;x<0

    (1)

    \displaystyle\ x-2>0\Rightarrow\frac{x}{x-2}<\frac{2x^2-4x+3}{x-2}\Rightarrow\ 2x^2-5x+3>0

    \Rightarrow\ (2x-3)(x-1)>0,\;\;x>2

    Both factors must be positive or both must be negative while x>2

    \displaystyle\Rightarrow\ x>\frac{3}{2},\;\;x>1,\;\;x>2\Rightarrow\ x>2

    Or

    \displaystyle\ x<\frac{3}{2},\;\;x<1,\;\;x>2 .... impossible

    (2)

    x-2<0,\;\;\;|x|=x

    We will be multiplying both sides by a negative value, so we reverse the inequality sign.

    x>2x^2-4x+3\Rightarrow\ (2x-3)(x-1)<0

    \displaystyle\ x>\frac{3}{2},\;\;x<1 ... impossible

    \displaystyle\ x<\frac{3}{2},\;\;x>1 ... a valid solution.


    (3)

    x<0\Rightarrow\ |x|=-x,\;\;x-2<0

    \displaystyle\Rightarrow\frac{-x}{x-2}<\frac{2x^2-4x+3}{x-2}\Rightarrow\ 2x^2-4x+3<-x\Rightarrow\ 2x^2+3x+3<0

    For this quadratic, f(0)=3 and it's roots are complex, since they are \displaystyle\frac{-3\pm\sqrt{3^2-4(2)3}}{2(2)}

    \Rightarrow\ 2x^2+3x+3>0

    The solution is

    x>2,\;\;\;\displaystyle\ 1<x<\frac{3}{2}
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  8. #8
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    Thanks to all your replies, I've managed to get the answer given by my text. Your approaches widened my perspective.

    If anyone would like to know how I got the answer, I'd be more than glad to do so here.
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