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Quadratic Equation

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  1. Dec 31, 2014 #1
    1. The problem statement, all variables and given/known data

    3x^2 + px + 3 = 0, p>0, one root is square of the other, then p=??



    2. Relevant equations
    sum of roots = -(coefficient of x)/(coefficient of x^2)

    product of roots= constant term/coeffecient of x^2

    3. The attempt at a solution
    ROot 1 = a
    root 2 = b
    b=a^2
    a.b= 3/3
    a^3=1
    a=1
    a^2=b=1
    a+b=-p/3=2
    p=-6

    Where did i go wrong??
     
  2. jcsd
  3. Dec 31, 2014 #2

    Ray Vickson

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    Nowhere: a = 1, p = -6 is the answer; it is the only real solution, but there are two other complex ones.

    However, I think you need more explanation. Why do you set a^3 = 1? Why can you set a+b=2? These are true, but you need to give reasons for saying them. I think a better presentation would be to say ##3 x^2 + px + 3 = 3(x-a)(x-a^2)## and just expand out the factorization on the right.
     
  4. Dec 31, 2014 #3

    Mark44

    Staff: Mentor

    I also got a = 1, p = -6, but there is a condition given that p > 0.
     
  5. Dec 31, 2014 #4

    Ray Vickson

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    OK: I missed that. If we allow complex roots (but insist on a real p) there is a value of p > 0 that works.

    When I said p = -6 is the only real solution, that was misleading: it is the only solution in which both p and and the roots are real. There are others with real p but complex roots.
     
    Last edited: Dec 31, 2014
  6. Dec 31, 2014 #5
    Someone said that I could use cube roots of unity?
    What exactly is that??
     
  7. Dec 31, 2014 #6

    Ray Vickson

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  8. Jan 2, 2015 #7

    lurflurf

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    There are two candidates for p one positive one negative, you found the wrong one.
    3x^2 + px + 3 = 0, p>0
    so the two roots are
    $$w=\dfrac{-p+\sqrt{p^2-36}}{6} \text{ and } w^2=\dfrac{-p-\sqrt{p^2-36}}{6}$$
    that is not the important part
    we know as you argued above
    w^3=1
    p=-3(w+w^2)
    can you use that to write p^2 in terms of p?
    After that solve for p.
     
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