# Given a cubic $f(x)$ with specified negative real roots $-a,-b,-c$, what happens when we search for solutions to $f(x)=d$?

Noting Roots of a Certain type of Cubic Equation, what if we have the following simpler form for real $d$:

$$(x+a)(x+b)(x+c)=d\tag{1}$$

(With $a,b,c\in \mathbb R^+$.) Clearly, depending on $d$, the number of real roots can be changed from $3$ to $1$ easily. Is it possible to get a simple form for the solution(s)?

I got as far as to see that real root $x=k$ is a solution to the following when $0\lt a\le b\le c$ and $d\in\mathbb R^+$:

$$k(k-a+b)(k-a+c)=d$$

Every time I attempt to run through the various mechanisms to solve the cubic with these conditions, the number of terms in each expression explodes in a way that makes the effort untenable. It almost seems like this equation should be analyzed in an entirely different way and that it would end up with a different but similar set of solutions relative to the standard set.

Any hints or solutions would be appreciated. Note that I am primarily interested in the real root(s) for $d\in\mathbb R^+$.

• I'm guessing you mean $a,b,c,d$ rational. If putting $d$ on the other side gives at least one rational root, you get a linear times a quadratic, finish with quadratic formula. If you get one real root but irrational, Cardano's formula works and can be untangled in the real numbers; meaning the real and imaginary parts of the cube roots involved can be found. If you arrive at three real irrational roots, you get nothing good, en.wikipedia.org/wiki/Casus_irreducibilis – Will Jagy Oct 3 '13 at 21:32
• @WillJagy: No, I mean generic real numbers... I'll check out the Casus irreducibilis that you mentioned. The intent is to consider continuously variable $d$ over the positive reals... – abiessu Oct 3 '13 at 21:33