<s>[INST] Finding the Sum of $a,b,c,d$ that satisfy the following conditions 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 $6$. 
 [/INST] 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 $a = -5, b = 6, d = 3$.
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