1
$\begingroup$

I would like to solve this polynomial of third order for its roots:

Reduce[{1 > a > 0, 1 > b > 0, 1 > c > 0, 1 > d > 0, 1 > f > 0, 1 > g > 0, (a (x + c + d) (x + f) + (b (x + d) + x (x + c + d)) (x + g + f)) == 0}, x]

Is it possible and if so is there a better way than reduce with the above assumptions?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$ 
FindInstance[{1 > a > 0, 1 > b > 0, 1 > c > 0, 1 > d > 0, 1 > f > 0, 1 > g > 0, (a (x + c + d) (x + f) + (b (x + d) + x (x + c + d)) (x + g + f)) == 0}, {a, b, c, d, f, g, x}] (* {{a -> 1/2, b -> 1/2, c -> 1/2, d -> 1/2, f -> 1/2, g -> 1/2, x -> 1/2 (-2 - Sqrt[2])}} *) a = b = c = d = f = g = 1/2; Reduce[(a (x + c + d) (x + f) + (b (x + d) + x (x + c + d)) (x + g + f)) == 0, x]

enter image description here

These are the only real roots, I think so.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much. I am interested in complex roots though, because I want to inverse transform this using the residue theorem. $\endgroup$ Commented Apr 1, 2016 at 8:08
  • $\begingroup$ @ Asking Questions You have an expression in x^3 and you have 3 roots (real). What do you want more? $\endgroup$ Commented Apr 1, 2016 at 9:05

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.