How to solve the following system of equations
$$ \begin{cases} a+c=12\\ b+ac+d=86\\ bc+ad=300\\ bd=625\\ \end{cases} $$
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$\begingroup$ 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? $\endgroup$– saulspatzFeb 25, 2018 at 3:15
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3$\begingroup$ What are the chances... $\endgroup$– dxivFeb 25, 2018 at 3:41
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$\begingroup$ 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. $\endgroup$– Claude LeiboviciFeb 25, 2018 at 6:07
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2 Answers
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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?
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$\begingroup$ 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
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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.
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$\begingroup$ You want $b,$ not $d$ in the denominator of the fraction. $\endgroup$ Feb 25, 2018 at 3:24
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$\begingroup$ From that onwards it become nightmare thanks for reply $\endgroup$ Feb 25, 2018 at 3:45
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$\begingroup$ Can we solve them analytically or have to opt for numerical methods $\endgroup$ Feb 25, 2018 at 3:51