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Roots of a quadratic equation.

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\alpha[/itex] and [itex]\alpha^{2}[/itex] are two roots of the equation x[itex]^{2}[/itex] -12x + k = 0

    Find 2 values for k.


    3. The attempt at a solution

    [itex]\alpha[/itex] + [itex]\alpha[/itex][itex]^{2}[/itex] = 12
    [itex]\alpha[/itex][itex]^{3}[/itex] = k

    I have no idea where to go from here. Any help appreciated.
     
    Last edited by a moderator: Dec 30, 2011
  2. jcsd
  3. Dec 30, 2011 #2

    Mark44

    Staff: Mentor

    Solve for α in the equation α2 + α - 12 = 0. Then substitute into your second equation to find k.
     
  4. Dec 30, 2011 #3
    a(1 + a) = 12
    a = 12 a = 11


    12[itex]^{2}[/itex] - 12(12) + k = 0
    k = 0

    11[itex]^{2}[/itex] -12(11) + k = 0
    k = 11

    Is this right, because the answers at the back of my book are k = -64 k = 27
     
  5. Dec 30, 2011 #4

    Mark44

    Staff: Mentor

    This is not how to solve a quadratic equation. Read what I said in post #2.
     
    Last edited: Dec 30, 2011
  6. Dec 30, 2011 #5
    I apologize both to you and myself for such an error in my judgement.

    Is there a rule as to when you are allowed to factorize via taking out what's common? Does the equation have to equal zero?
     
  7. Dec 30, 2011 #6

    Mark44

    Staff: Mentor

    Let me be a bit pedantic here by saying that an equation does not equal a number. An expression can be equal to a number, but an equation already has an '=' mark in it, so it would be meaningless to say that an equation is equal to zero or any other number.

    What you might be alluding to is a theorem that says if the product of two numbers is 0, then one or the other of the numbers, or both, must be 0.

    What you had was a(a + 1) = 12, from which you concluded that a = 12 or a = 11. This doesn't work at all. There are lots of ways that two numbers can multiply to 12, such as 1*12, 2*6, 3*4, (1/3)*36, -1*(-12), etc., etc. If you had checked your work, by replacing a with 12, you would have seen that 12(13) ≠ 12. Similarly, if a = 11, you would have seen that 11(12) ≠ 12.

    But, by writing the equation as a2 + 1 - 12 = 0, and then factoring or using the Quadratic Formula, you should come out with the values that are solutions to the equation.
     
  8. Dec 30, 2011 #7
    Ah ok, I see. Thanks for the clarification!
     
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