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Thread: Solving an equation by substitution

  1. #1
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    Solving an equation by substitution

    Hello, I've come across a problem in my textbook that really baffles me.

    Task: to use substitution $\displaystyle u=x+1/x$ to solve the equation for x:

    $\displaystyle x^2+3x-2+3/x+1/x^2=0$

    It was easy to make the quadratic equation,

    $\displaystyle u^2+3u-2=0$

    $\displaystyle u1= (-3+sqrt(17))/2$
    $\displaystyle u2=(-3-sqrt(17))/2$

    but solving for x was a bit "ugly" (quoted because math is NOT ugly )

    $\displaystyle x+1/x=(-3+sqrt(17))/2$
    $\displaystyle x+1/x=(-3-sqrt(17))/2$

    I've spent quite a time over this, any pointers?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by VImtermute View Post
    Hello, I've come across a problem in my textbook that really baffles me.

    Task: to use substitution $\displaystyle u=x+1/x$ to solve the equation for x:

    $\displaystyle x^2+3x-2+3/x+1/x^2=0$

    It was easy to make the quadratic equation,

    $\displaystyle u^2+3u-2=0$

    $\displaystyle u1= (-3+sqrt(17))/2$
    $\displaystyle u2=(-3-sqrt(17))/2$

    but solving for x was a bit "ugly" (quoted because math is NOT ugly )

    $\displaystyle x+1/x=(-3+sqrt(17))/2$
    $\displaystyle x+1/x=(-3-sqrt(17))/2$

    I've spent quite a time over this, any pointers?
    Multiply through by $\displaystyle x$ and rearrange, to get quadratics.

    RonL
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  3. #3
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    Thanks for the reply CaptainBlack.
    That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

    P.S.
    The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$
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  4. #4
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    Quote Originally Posted by VImtermute View Post
    Thanks for the reply CaptainBlack.
    That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

    P.S.
    The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$
    I don't like the look of your roots to that quadratic in $\displaystyle u$.

    I think that:

    $\displaystyle
    x^2+3x-2+3/x+1/x^2=0
    $

    gives:

    $\displaystyle
    u^2+3u-4=0
    $

    under the proposed substitution, which has roots $\displaystyle (u=-4) \vee (u=1)$

    RonL
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  5. #5
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    !!!

    CaptainBlack, you're absolutely right. I always forget the rules for squaring polynomials

    Thank you very much.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by VImtermute View Post
    Thanks for the reply CaptainBlack.
    That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

    P.S.
    The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$
    First off: There are four solutions to this equation. Two complex and two real. You have listed only the two real solutions.

    Second: Your quadratic in u is incorrect.

    $\displaystyle x^2+3x-2+3/x+1/x^2=0$

    $\displaystyle u = x + \frac{1}{x}$
    so
    $\displaystyle u^2 = x^2 + 2 \frac{1}{x^2}$

    So we wish to have
    $\displaystyle au^2 + bu + c = ax^2 + bx + c + \frac{b}{x} + \frac{b^2}{x^2} = x^2 + 3x - 2 + \frac{3}{x} + \frac{1}{x^2}$

    So
    a = 1
    b = 3
    c = -4

    Thus
    $\displaystyle u^2 + 3u - 4 = 0$

    $\displaystyle (u + 4)(u - 1) = 0$

    So u = -4 or u = 1.

    Thus
    $\displaystyle x + \frac{1}{x} = -4$

    $\displaystyle x^2 + 4x + 1 = 0$ ==> $\displaystyle x = -2 \pm \sqrt{3}$

    and
    $\displaystyle x + \frac{1}{x} = 1$

    $\displaystyle x^2 - x + 1 = 0$ ==> $\displaystyle x = \frac{1 \pm \sqrt{3}}{2}$

    -Dan
    (Rats! FOILed again. )
    Last edited by topsquark; Jun 15th 2007 at 06:53 AM. Reason: I was beaten!!
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