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Proof of x^3+a^2x+b^2=0 roots question

  1. Mar 20, 2012 #1
    I want to prove x^3 + a^2x + b^2 = 0 has one negative and two imaginary roots if b \not= 0.

    I know that it cannot have any positive real roots because a > 0 and b > 0 will always be the case.

    I believe I can prove this using Descarte's Rule of Signs (which is in the same chapter of this problem).

    Using the rule of Signs I get f(-x) = -x^3 -a^2x + b^2 = 0 which implies one negative root. Thus we conclude that we have two imaginary roots since we have to have a total of 3 roots and none can be positive.

    However, I do am not sure if this is exactly what is meant when it says to prove it. Does this suffice?

    Also, how would I begin to prove this algebraically?
     
    Last edited: Mar 20, 2012
  2. jcsd
  3. Mar 20, 2012 #2

    mathman

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    You need to rule out the possibility of 3 real negative roots. Hint: sign of constant term.
     
  4. Mar 20, 2012 #3
    Hey,

    You have done well. :-)


    The total number of roots (real + imaginary)should be 3.
    And imaginary roots always occur in pairs (Why?)

    So either the number of real roots is 1 or 3.

    Remember the sign rule on f(x) tells you about the MAXIMUM no. of possible positive roots.


    In your first step using the sign rule you derived that maximum no. of positive roots is 0.
    So that means no positive roots exist.

    Now using sign rule on f(-x)
    You got maximum no. Of negative roots is 1.

    So no. negative roots can be both 0 and 1.
    But since we know that the equation should have atleast one real root,
    You now know that number of negative roots is 1.

    This is valid as long as both a and b are not zero.

    You will have to check again for the case where a is 0 to get your entire answer
     
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