Let $a,b,c,d$ represent $4$ different non-zero integers such that the absolute value of each integer is less than $11$. If $c$ and $d$ are the solutions for $x$ of $x^2+ax+b=0$ and if $a$ and $b$ are the solutions for $x$ of $2x^2-cx-20d=0$, find the value of $a+b+c+d$.
I know the answer to this problem should be $6$.
From Vieta's formulas we know that: $$ \left\{ \begin{array}{rrrrrr} c+d &= &-\dfrac{a}{1}\\ cd &= &\dfrac{b}{1}\\ a+b &= &-\dfrac{-c}{2}\\ ab &= &\dfrac{-20d}{2}\\ \end{array} \right. $$ Then $$ \left\{ \begin{array}{rrrrrr} a &= &-c -d\\ b &= &cd\\ -c-d+cd &= &\dfrac{c}{2}\\ -(c+d)cd &= &-10d\\ \end{array} \right. $$ Next $$ \left\{ \begin{array}{rrrrrr} d &= &\dfrac{3c}{2(c-1)}\\ c^2 + cd &= &10\\ \end{array} \right. $$ Therefore $$ c^2 +c \dfrac{3c}{2(c-1)} =10 $$ Thus $$ 2c^3 + c^2 -20c + 20 = 0 $$ After solving this equation we will get $c=2$ (and two other solutions which are not integers).
So $a = -5, b = 6, d = 3$.
