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Finding the roots of a fourth degree polynomial

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the roots of x^4 - 6x^2 - 2

    2. Relevant equations


    3. The attempt at a solution
    So my first observation is that this polynomial is irreducible by Eisenstein criterion with p=2. If I substitute y=x^2 then this polynomial becomes a quadratic, and I can apply the quadratic equation to get two solutions for y. Then could I take the +/- of the square root of these solutions to get the roots in terms of x? If so, is there a simpler way to find these roots? Thanks PF!
     
  2. jcsd
  3. Nov 16, 2016 #2

    fresh_42

    Staff: Mentor

    I find this method pretty fast. The check whether it gives the correct answer has been longer than to write down the solution. And the square roots have to appear somewhere. What do you have in mind?
     
  4. Nov 17, 2016 #3
    What do you mean the square roots have to appear somewhere? Yes, the check whether it gives a correct answer seems to be longer, but I could use Wolfram. I don't have any other methods in mind to solve the roots of this polynomial, I was just wondering if there was a more standard/methodical method I could use.
     
  5. Nov 17, 2016 #4

    fresh_42

    Staff: Mentor

    Well, ##y^2 + px +q = 0## lets us directly write ##y_{1,2}=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2 - q}## which yields ##x_{1,2,3,4} = \pm (\sqrt{3 \pm \sqrt{11}})##.
    I meant ##\sqrt{11}## and ##\sqrt{3\pm \sqrt{11}}## have to be found somehow. So any other method has to output them and I cannot think of any faster method.
     
  6. Nov 17, 2016 #5
    Are these not all four roots we were looking for? With +/- at the beginning of course.
     
  7. Nov 17, 2016 #6

    fresh_42

    Staff: Mentor

    Yes, and that's why I found it rather short and fast.
     
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