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Math Help - Help finding roots

  1. #1
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    Help finding roots

    Hi, I need to solve for x in the following two equations:

    (x-1/x)^2 - 77/12 (x-1/x) + 10 = 0

    [I expanded out to get (x^2 + 1/x^2 - 77x/12 + 77/12x + 8). Then I used the factor theorem to plug in possible x's, and found that (x-4) is a factor. However, I don't know where to go from there, or if what I've done so far is correct.]

    and

    (3x-5)(3x+1)^2(3x+7) + 68 = 0

    Thank you for any help!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    Hi, I need to solve for x in the following two equations:

    (x-1/x)^2 - 77/12 (x-1/x) + 10 = 0

    [I expanded out to get (x^2 + 1/x^2 - 77x/12 + 77/12x + 8). Then I used the factor theorem to plug in possible x's, and found that (x-4) is a factor. However, I don't know where to go from there, or if what I've done so far is correct.]

    and

    (3x-5)(3x+1)^2(3x+7) + 68 = 0

    Thank you for any help!
    don't expand! it is quadratic in (x - 1)/x

    replace (x - 1)/x with another variable, say y, and you will see that

    y^2 - (77/12)y + 10 = 0

    solve for the roots of that equation, when you get them, replace y with (x - 1)/x and solve for x


    EDIT: Oh, it was x - 1/x not (x - 1)/x. that does not change the process though

    EDIT 2: by the way, you expanded incorrectly

    EDIT 3: I must be tired. the expansion was correct, so scratch out EDIT 2. I still think my original plan is best though. Yes you have to do the quadratic formula 2 or 3 times (2 if you use Plato's method), but to that's easier than working with the quartic equation you would get if you continued down the path you were going
    Last edited by Jhevon; July 8th 2007 at 10:16 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Let's use Jhevon's method.

     y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

     y = \frac{(77/12) \pm \sqrt{(-77/12)^2 - 4(1)(10)}}{2}

     y = \frac{(77/12) \pm \sqrt{1,1736}}{2}

     y = 3,795 \ or \ y = 2,622
    incomplete, remember we want the x-values. so assuming what you did is correct (i have no time to check it) you now replace each y with x - 1/x and solve for x to find the corresponding x-values
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by janvdl View Post
    Let's use Jhevon's method... and the mathematician's best friend, the quadratic formula

     y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

     y = \frac{(77/12) \pm \sqrt{(-77/12)^2 - 4(1)(10)}}{2}

     y = \frac{(77/12) \pm \sqrt{1,1736}}{2}

     y = 3,795 \ or \ y = 2,622


     x - \frac{1}{x} = 3,795

     x^2 - 3,975x - 1 = 0

    Use the quadratic formula again.

    Then you'll see that x = 3.302 or
    x = -0.302



     x - \frac{1}{x} = 2,622

     x^2 - 2,622x - 1 = 0

    Use the quadratic formula AGAIN.

    Then you'll get x = 2.414
    x = -0.414
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  5. #5
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    y = \left( {x - \frac{1}{x}} \right)\quad  \Rightarrow \quad 12y^2  - 77y + 120 = 0\quad  \Rightarrow \quad \left( {3y - 8} \right)\left( {4y - 15} \right) = 0

    Now you solve for x:
     \left( {x - \frac{1}{x}} \right) = \frac{8}{3}\quad \& \quad \left( {x - \frac{1}{x}} \right) = \frac{{15}}{4}
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