Roots of a polynomial

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  1. Zishi's Avatar
    1 21-06-2012  04:43
    • Peer Of The TSR Realm
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    Roots of a polynomial
    My problem is with this question:


    The two roots which I get are \dfrac{-p + \sqrt{p^2-108}}{6} and \dfrac{-p - \sqrt{p^2-108}}{6}. Because squaring a number always gives you a positive number, so the square of \dfrac{-p - \sqrt{p^2-108}}{6}(as it is a negative number) should be equal to \dfrac{-p + \sqrt{p^2-108}}{6}.

    \Rightarrow \left(\dfrac{-p - \sqrt{p^2-108}}{6}\right)^2 = \dfrac{-p + \sqrt{p^2-108}}{6}

    My problem is that when I simplify this equation, I get a very complex expression which is difficult to solve further. Does anyone has another easy way to do this question?

    Any help would be appreciated!
    Last edited by Zishi; 21-06-2012 at 06:30.
  2. Astronomical's Avatar
    2 21-06-2012  04:53
    • Overlord in Training
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    Re: Roots of a polynomial
    What expression do you get?
  3. notnek's Avatar
    3 21-06-2012  04:58
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    Re: Roots of a polynomial
    If the roots of a quadratic equation ax^2+bx+c=0 are x_1 and x_2 then

    x_1+x_2=-\frac{b}{a}

    x_1 x_2 = \frac{c}{a}

    Do these help?

    If you're not familiar with them then you could call one of the roots r and the other r^2 then compare coefficients with the expanded form of

    (x-r)(x-r^2)=0
  4. ghostwalker's Avatar
    4 21-06-2012  05:02
    • Outcast of Imrryr
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    Re: Roots of a polynomial
    I don't see how p can be positive, but since I've had little sleep for the last three nights, my brain's probably on autopilot.
    Last edited by ghostwalker; 21-06-2012 at 05:48.
  5. raheem94's Avatar
    5 21-06-2012  05:19
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    Re: Roots of a polynomial
    (Original post by ghostwalker)
    I don't see how p can be positive, but since I've had little sleep for the last three nights, my brain's probably on autopilot.
    The question is may be flawed.

    The correct answer is probably p=-6, but why does the question says p>0 and then includes a negative value of p is confusing.
  6. Zishi's Avatar
    6 21-06-2012  05:31
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    Re: Roots of a polynomial
    (Original post by Astronomical)
    What expression do you get?
    p^2 + 2p\sqrt{p^2-108} + p^2 - 108 = -6p + 6\sqrt{p^2 - 108}

    (Original post by notnek)
    If the roots of a quadratic equation ax^2+bx+c=0 are x_1 and x_2 then

    x_1+x_2=-\frac{b}{a}

    x_1 x_2 = \frac{c}{a}

    Do these help?

    If you're not familiar with them then you could call one of the roots r and the other r^2 then compare coefficients with the expanded form of

    (x-r)(x-r^2)=0
    I didn't know about first rule, using that gives r^3 = 1, which in turn gives the value of -6. But the answer key to this question gives the value of p as 3, i.e (c)

    Using the second rule - The full form of the equation you've given is x^2 - rx - r^2x +r^3. Multiplying this by 3 and comparing the coefficients, this gives r^3 = 1 and again the value of p as -6.

    I think this means that the question is flawed, as pointed out by ghostwalker and raheem. :dontknow:
  7. ghostwalker's Avatar
    7 21-06-2012  05:33
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    Re: Roots of a polynomial
    (Original post by raheem94)
    ...
    Yes, I'm brain dead. I was ignoring the possibility that the roots are complex.

    3 does work.
    Last edited by ghostwalker; 21-06-2012 at 05:38.
  8. raheem94's Avatar
    8 21-06-2012  05:45
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    Re: Roots of a polynomial
    (Original post by Zishi)
    p^2 + 2p\sqrt{p^2-108} + p^2 - 108 = -6p + 6\sqrt{p^2 - 108}



    I didn't know about first rule, using that gives r^3 = 1, which in turn gives the value of -6. But the answer key to this question gives the value of p as 3, i.e (c)

    Using the second rule - The full form of the equation you've given is x^2 - rx - r^2x +r^3. Multiplying this by 3 and comparing the coefficients, this gives r^3 = 1 and again the value of p as -6.

    I think this means that the question is flawed, as pointed out by ghostwalker and raheem. :dontknow:
    http://www.wolframalpha.com/input/?i...B+3+x+%2B3%3D0

    As ghostwalker says it does work.

    The complex solutions are \displaystyle x = - \sqrt[3]{-1} \text{ and } x = (-1)^{\frac23} when p=3.
  9. ghostwalker's Avatar
    9 21-06-2012  05:47
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    Re: Roots of a polynomial
    OK,

    Since r^3=1

    then r=1\text{ or }\omega\text{ or }\omega^2

    We reject r=1 as this gives a negative value for p, of -6.

    So, the roots of our equation are then \omega\text{ and }\omega^2

    and p = -3(\text{sum of roots})=-3(\omega+\omega^2)=-3(-1)=3

    Multiplying by 3 as that's the coeff of x^2.
    Last edited by ghostwalker; 21-06-2012 at 05:59. Reason: Add a detail.
  10. Zishi's Avatar
    10 21-06-2012  05:47
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    Re: Roots of a polynomial
    (Original post by ghostwalker)
    Yes, I'm brain dead. I was ignoring the possibility that the roots are complex.

    3 does work.
    Hmm. Taking r = e^{\frac{2\pi}{3}i} gives p as 3. Many thanks. Btw this question has to be done without the use of a calculator - is there any simple way to find values of \cos \frac{2\pi}{3} and \sin \frac{2\pi}{3}? Or do I have to remember their values?
  11. ghostwalker's Avatar
    11 21-06-2012  05:52
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    Re: Roots of a polynomial
    (Original post by Zishi)
    Hmm. Taking r = e^{\frac{2\pi}{3}i} gives p as 3. Many thanks. Btw this question has to be done without the use of a calculator - is there any simple way to find values of \cos \frac{2\pi}{3} and \sin \frac{2\pi}{3}? Or do I have to remember their values?
    If you recall the shape of the sin/cos curves, then

    \cos \frac{2\pi}{3}=-\cos \frac{\pi}{3}

    \sin \frac{2\pi}{3}=\sin \frac{\pi}{3}

    PS: It's probably easier to work with the cube roots of unity, if you've covered them.
    Last edited by ghostwalker; 21-06-2012 at 05:55.
  12. Zishi's Avatar
    12 21-06-2012  06:08
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    Re: Roots of a polynomial
    (Original post by ghostwalker)
    If you recall the shape of the sin/cos curves, then

    \cos \frac{2\pi}{3}=-\cos \frac{\pi}{3}

    \sin \frac{2\pi}{3}=\sin \frac{\pi}{3}

    PS: It's probably easier to work with the cube roots of unity, if you've covered them.
    Hmm, thanks again. (PRSOM)
  13. Astronomical's Avatar
    13 21-06-2012  12:35
    • Overlord in Training
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    Re: Roots of a polynomial
    Haven't tried solving it myself, but out of curiosity I plugged it into WolframAlpha and apparently p is roughly -10.

    http://www.wolframalpha.com/input/?i...%7D%7D%7B6%7D+
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