Say I have an equation of the form

$$u s + \frac{1}{v} + \frac{1}{p s + q} = 0$$

(or any form that can be written as a standard quadratic, really, the above form is just an example; they'll all be equal to 0 from the start, however) and I want it in the form

$$ s^2 + b s + c = 0 $$ i.e. the coefficient on $s^2$ is 1.

How do I do this using Mathematica? More specifically, I'm interested in finding $b$ and $c$, but if the equation is written in this form, grouped properly, doing that is very easy by eye.


2 Answers 2


You can combine the terms to a common denominator using Together:

Together[u s + 1/v + 1/(p s + q)]

$\displaystyle\frac{p u v s^2 +p s+q s u v+q+v}{v (p s+q)}$

Now, assuming $v (p s+q)\neq 0$, you can get your coefficients $c$ and $b$ respectively as:

Most@#/Last@# &@CoefficientList[Numerator[Together[u s + 1/v + 1/(p s + q)]], s]
(* {(q + v)/(p u v), (p + q u v)/(p u v)} *)
  • $\begingroup$ It's hard to pick an answer when both answers work great and were posted less than two minutes apart! In the end, I went by time posted. Thanks to both of you, though. $\endgroup$
    – exscape
    Commented May 30, 2012 at 6:55

Assuming that your equations are quadratic in $s$, here's one way:

(Collect[#/Coefficient[#, s, 2], s]) &[Numerator[Together[u s + 1/v + 1/(p s + q)]]]
s^2 + (q + v)/(p u v) + (s (p + q u v))/(p u v)

If it's just the coefficients you want:

(CoefficientList[#/Coefficient[#, s, 2], s]) &[Numerator[Together[u s + 1/v + 1/(p s + q)]]]
{1/(p u) + q/(p u v), q/p + 1/(u v), 1}

Note the convention of putting the constant term first and leading coefficient last.


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