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Math Help - Inequality

  1. #1
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    Inequality

    Hey guys I'm confused whether I doing these right, I used 2 different methods to solve this problem, and they seem to be giving different answers

    \text{solve}  \dfrac{1}{x-3}> 4

    1st method:

     \dfrac{1}{x-3}> 4

    \dfrac{1}{(x-3)^2}>16

    1>16(x-3)^2

    1>16x^2-96x+144

    16x^2-96x+143<0

    x=\dfrac{96\pm\sqrt{96^2-4(16)(143)}}{32}

    \dfrac{88}{32}<x<\dfrac{13}{4}

    2nd method

     \dfrac{1}{x-3}> 4

    \dfrac{1}{x-3}-4>0

    \text{multiply both sides by}\hspace{5mm}(x-3)^2

    (x-3)-4(x-3)^2>0

    (x-3)(1-4(x-3))>0

    (x-3)(13-4x)>0

    3<x<\dfrac{13}{4}



    they give different answers...? Pls let me know if i've done anything wrong

    Thanks
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  2. #2
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    Whenever you square an equation or inequation to aid in solving it, you are likely to bring in extraneous solutions.

    What I would do is invert both sides, noting that inverting both sides of an inequation changes the direction of the inequality sign. The rest should be obvious.
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  3. #3
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    First we need to get a common denominator on both sides

    \displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}

    simplify to get

    \displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}

    So the two critical numbers(where the inequality can chage) are \displaystyle x=3,\quad x=\frac{13}{4}

    If you test some points you will get

    \displaystyle \left(3,\frac{13}{4}\right)
    Last edited by TheEmptySet; April 9th 2011 at 07:38 AM.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    First we need to get a common denominator on both sides

    \displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}

    simplify to get

    \displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}

    So the two critical numbers(where the inequality can chage) are \displaystyle x=3,quad x=\frac{13}{4}

    If you test some points you will get

    \displaystyle (-\infty,3) \cup \left(\frac{13}{4},\infty \right)
    Sorry but for any \displaystyle x in \displaystyle (-\infty, 3), \displaystyle \frac{1}{x - 3} gives something negative, which is clearly NOT greater than \displaystyle 4...
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  5. #5
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    The way I would have done it is this: \frac{1}{x-3}> 4 is the same as 0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals (-\infty, 3), (3, 13/4), and (13/4, \infty). If x= 0, which is less than 3, \frac{1}{0-3}= -\frac{1}{3}< 4. If x= 4, which is larger than 13/4, \frac{1}{4-3}=  1< 4. If x= 3.1, which is between 3 and \frac{13}{4}= 3\frac{1}{4}, \frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4.

    Therefore, the inequality is true on the interval \left(3, \frac{13}{4}\right).
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by Prove It View Post
    Sorry but for any \displaystyle x in \displaystyle (-\infty, 3), \displaystyle \frac{1}{x - 3} gives something negative, which is clearly NOT greater than \displaystyle 4...
    You are correct I cant add this morning! thanks
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  7. #7
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    "If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.

    The second method is wrong. You don't know if \frac{(x-3)^2}{x-3}=x-3 or 3-x.

    First method is wrong too. You can square only after you say that x must be bigger then 3.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    The way I would have done it is this: \frac{1}{x-3}> 4 is the same as 0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals (-\infty, 3), (3, 13/4), and (13/4, \infty). If x= 0, which is less than 3, \frac{1}{0-3}= -\frac{1}{3}< 4. If x= 4, which is larger than 13/4, \frac{1}{4-3}=  1< 4. If x= 3.1, which is between 3 and \frac{13}{4}= 3\frac{1}{4}, \frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4.

    Therefore, the inequality is true on the interval \left(3, \frac{13}{4}\right).
    Actually \displaystyle \frac{1}{x - 3} > 4 is the same as \displaystyle 0 > 4 - \frac{1}{x - 3}...
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  9. #9
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    A way this kind of inequalities can be solved:

    \frac{1}{x-3}>4 \Leftrightarrow  4-\frac{1}{x-3}<0 \Leftrightarrow  \frac{4x-13}{x-3}<0



    You want that fraction be negativ, so x \in \left ( 3, \frac{13}{4} \right ).
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  10. #10
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    Re: Inequality

    Quote Originally Posted by veileen View Post
    The second method is wrong. You don't know if \frac{(x-3)^2}{x-3}=x-3 or 3-x.
    sorry for bringing up an old thread, but I thought about this inequality thing again, looking over what you wrote isn't \frac{(x-3)^2}{x-3}=x-3 always?, isn't this true for all real numbers x, for example sub in x=5 then the answer would be 2, which is 5-3, if subbing in lets say a negative number -1, then it would be 16/-4=-1-3 which is true again, the same is true for x=0 or a positive number smaller than 3, eg 1
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  11. #11
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    Re: Inequality

    The statement:
    \frac{(x-3)^2}{x-3}=3-x is not true because, that would mean (3-x)\cdot(x-3)=(x-3)^2 \Leftrightarrow -x^2+6x-9=x^2-6x+9 which is not true for all x.
    Last edited by Siron; August 14th 2011 at 05:16 AM.
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  12. #12
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    Re: Inequality

    Quote Originally Posted by veileen View Post
    "If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.
    There are a lot of things that anybody can do- that doesn't mean it is wrong.

    The only places an inequality can change form ">" to "<" is where the function is 0 or is not continuous. For fractions that is where the numerator is 0 or where the denominator is 0. Once you have found those points, checking one point in each interval tells you which. What TheEmptySet said was perfectly valid.
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