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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.

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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. 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
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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 –  belisarius May 5 '13 at 19:05
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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
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I would leave this as a comment if I could...

You can solve these analytically using some complex analysis and trigonometry as per this link.

Also Tristan Needham's excellent Visual Complex Analysis deals with solving cubics in depth. I know its not Mathematica *per se*, but should be usable along side of it. The trigonometric solution might be a little simpler than what is achieved by the mess that is:

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