How to solve the following system of equations

$$ \begin{cases} a+c=12\\ b+ac+d=86\\ bc+ad=300\\ bd=625\\ \end{cases} $$

  • $\begingroup$ Welcome to MSE. You'll get a lot more help and a lot fewer down votes if you provide some context for your questions. What have yo done so far, and where are you stuck? $\endgroup$
    – saulspatz
    Feb 25, 2018 at 3:15
  • 3
    $\begingroup$ What are the chances... $\endgroup$
    – dxiv
    Feb 25, 2018 at 3:41
  • $\begingroup$ If $a,b,c,d$ are real numbers, there will be two sets of solutions. If they may be complex, there will be probably six. $\endgroup$ Feb 25, 2018 at 6:07

2 Answers 2


This system of equations corresponds to a factorization of $$f(x)=x^4+12x^3+86x^2+300x+625$$ into two quadratics as $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ The roots of $f(x)$ are $-3-4i$ and $-3+4i$, each having multiplicity $2$. So what should the quadratic factors of $f(x)$ look like?

  • $\begingroup$ Thanks for reply these equations are obtained by comparison. I have calculated by matlab. But need to know how to solve these set of equations. By Substitution it gives 4 order equation. $\endgroup$ Feb 25, 2018 at 3:47

Well, $b=\cfrac {625}{d}$ from your last equation, and $a=-c+12$ from your first equation.

Substitute $b$ and $a$ into your second equation to get $\cfrac{625}{d}+(-c+12)(c)+d=86$

Do the same for your third equation.

  • $\begingroup$ You want $b,$ not $d$ in the denominator of the fraction. $\endgroup$
    – saulspatz
    Feb 25, 2018 at 3:24
  • $\begingroup$ From that onwards it become nightmare thanks for reply $\endgroup$ Feb 25, 2018 at 3:45
  • $\begingroup$ Can we solve them analytically or have to opt for numerical methods $\endgroup$ Feb 25, 2018 at 3:51

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