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Australian HSC mathematics extension 2 exam Polynomial

  1. Oct 27, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    This problem is from the Australian HSC mathematics extension 2 exam. Q6b)

    It states:

    b) Let [tex]P(x)=x^3+qx^2+qx+1[/tex], where q is real. One zero of P(x) is -1

    i) Show that if [itex]\alpha[/itex] is a zero of P(x) then [tex]\frac{1}{\alpha}[/tex] is a zero of P(x)

    ii) Suppose that [itex]\alpha[/itex] is a zero of P(x) and [itex]\alpha[/itex] is not real.
    (1) show that [tex]|\alpha|=1[/tex]
    (2) show that [tex]Re(\alpha)=\frac{1-q}{2}[/tex]


    3. The attempt at a solution

    i) I was able to answer this one

    ii) (1) I was uncertain about this one so I fudged the answer, but for a quick and unsatisfactory answer, I went with the idea that if [itex]\alpha[/itex] and [itex]1/\alpha[/itex] are complex roots to a real polynomial, then the roots must occur in conjugate pairs and thus have the same modulus. Actually... also

    [tex]P(x)=x^3+qx^2+qx+1\equiv (x+1)(x^2+(q-1)x+1)[/tex]

    The quadratic factor: [tex]x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}[/tex]
    Thought I had it... guess not...

    ii) (2) I scabbed the answer from the quadratic (without really understanding why it works) that [tex]Re(z)=Re(\alpha)=\frac{1-q}{2}[/tex]
    This is the part without the surd. Also, unsure of a correct approach to the answer.

    Help for these 2 parts would be greatful.
     
  2. jcsd
  3. Oct 27, 2009 #2

    Office_Shredder

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    Re: Polynomial

    For part ii, the idea that they have to come in conjugate pairs sounds good to me. What did you find wrong with it?

    For the last part notice [tex]
    x=\frac{1-q\pm\sqrt{(q-3)(q+1)}}{2}
    [/tex] is real unless the part inside the square root is negative. If that's negative, what's the real part of the number?
     
  4. Oct 27, 2009 #3

    Mentallic

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    Re: Polynomial

    for ii) I thought that while I was able to show that[tex]|\alpha|=|\frac{1}{\alpha}|[/tex], I wasn't able to show they're both equal to 1. But now that I look at it again, for a complex number [itex]\alpha[/itex] and its reciprocal to be equal, they must have a mod of 1.

    for iii) well yes the real part is ofcourse obvious, but this assumes that [tex](q-3)(q+1)<0[/tex] thus [tex]-1<q<3[/tex]
    How do I explain/show that q is in fact in this range? Or can I just assume it because the question specifically stated that the roots are imaginary?
     
  5. Oct 27, 2009 #4

    Office_Shredder

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    Re: Polynomial

    If it's not in that range, then you've found all the roots, and they are real. But one of them has to be imaginary, so that's a contradiction
     
  6. Oct 27, 2009 #5

    Mentallic

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    Re: Polynomial

    I understand now. Thanks for your help Office_Shredder :smile:
     
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