1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cubic equation roots.

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data
    The equation
    x3 − 3x2 + px + 4 = 0,
    where p is a constant, has roots α −β , α and α + β , where β > 0.
    (a) Find the values of α and β .
    (b) Find the value of p.
    how do i start off? all i know is that sigma a= -b/a and ab= c/a and ab(gamma) = -d/a .
    Would this be one of the ways to do it: [x-(a-β)] [x-(a+β)] (x-a).

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 7, 2010 #2


    User Avatar
    Homework Helper

    You know that ∑α=-b/a, so use that to find α. ∑αβ=c/a and ∑αβγ=-d/a.

    So start off using the first method. Sum of all of the roots and equate it to 3. Do a similar exercise for the product of the roots.
  4. Mar 7, 2010 #3
    Oo is this how it works: ∑α=-b/a => -(-3)/1 = 3.
    to find b: ∑ab=c/a
    = ∑a.∑b= c/a.
    3∑b=px ... b=px/3. <--- sorry for such a weird presentation. Not sure :(.
  5. Mar 7, 2010 #4


    User Avatar
    Homework Helper

    Yes, but ∑α is the sum of the roots, your roots are α−β , α and α + β, what are the sum of the roots?
  6. Mar 7, 2010 #5
  7. Mar 7, 2010 #6


    User Avatar
    Homework Helper

    Right yes good 3α.

    So ∑α=3α=3, what is α then?
  8. Mar 8, 2010 #7
    Oh! its one.. so to find out constant b would i mutiply out my roots:when a=1
    (1-b) (1+b) (1) = i think its wrong =/ .
  9. Mar 8, 2010 #8


    User Avatar
    Science Advisor

    Another way to do this would be to set
    [tex](x- \alpha- \beta)(x- \alpha)(x- \alpha+ \beta)= x^2- 3x^2+ px+ 4[/tex]
    multiply it out and set corresponding coefficients equal. That gives you three equations for [itex]\alpha[/itex], [itex]\beta[/itex], and p.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook