2014
Problem - 3320
Find one real solution $(a, b, c, d)$ to the following system:
$$
\left\{
\begin{array}{rcl}
a+b+c+d&=&-2\\
ab+ac+ad+bc+bd+cd&=&-3\\
abc+abd+acd+bcd&=&4\\
abcd&=&3
\end{array}
\right.
$$
By Vieta's theorem, $a, b, c$, and $d$ are the four roots of the following equation:
$$x^4 +2x^3 -3x^2 -4x +3=0$$
In order to solve this equation, we can cancel odd powers by substituting $x=t-\frac{1}{2}$ which leads to: $$16t^4 - 72t^2 + 65=0 \implies t^2_{1,2} = \frac{5}{4}, \frac{14}{4}$$
Hence we conclude one solution is $$\Big(\frac{\sqrt{5}-1}{2}, \frac{-\sqrt{5}-1}{2},\frac{\sqrt{13}-1}{2},\frac{-\sqrt{13}-1}{2}\Big)$$
Note: solve high degree equation using substitution is discussed in the book %%HREF%%Competition Algebra%%Home/books/110-competition-algebra%%.