# solving a cubic equation

I need to find the minimum $r$ and the maximum $k$ of the following cubic equation for which there does not exist three distinct real roots.

$rx^3-rkx^2+(r+k)x-rk=0$.

Is it possible to find such $r$ and $k$ analytically? Or if you can provide me help using mathematica, that would be fine too.Thanks.

• You could try interpreting the output of Reduce[Discriminant[r x^3 - r k x^2 + (r + k) x - r k, x] == 0, {r, k}, Reals] // FullSimplify. – J. M.'s ennui May 5 '13 at 18:19
• Maybe you could tell us the approach(es) you've tried? – bill s May 5 '13 at 18:20
• The result looks really messy. I could not interpret it. Can you give me the solution? – upaudel May 5 '13 at 18:27
• I upvoted your question, and then saw your comment "The result looks really messy. I could not interpret it. Can you give me the solution?" Please be aware that if you can't interpret your own problem results, the problem is (perhaps) above your abilities – Dr. belisarius May 5 '13 at 19:05
• You want the cubic polynomial's discriminant to be non-positive for it not to have 3 distinct real roots. If you look at the results of what @J.M. suggested, you should at least be able to handle readily some of the boundary cases that arise. – murray May 5 '13 at 19:59

Solve[r*x^3 - r*k*x^2 + (r + k)*x - r*k == 0, x]

• I agree that it should probably be a comment but I couldn't help upvoting, as I think it's an interesting point. For specific irreducible cubics, you can often (always??) express the roots in this form using ComplexExpand. For example: ComplexExpand[Re[z /. Solve[z^3 - 3 z - 1 == 0, z]]]. In this case, the Re simply removes the imaginary parts, which we know to be zero anyway. – Mark McClure May 6 '13 at 13:06