Vieta's Formula

Vieta's Formula
The equations  $x^3 + 5x^2 + px + q = 0$ and $x^3 + 7x^2 + px + r = 0$ have two roots in common.
If the third root of each equation is represented by $x_1$ and $x_2$ respectively,
compute the ordered pair $(x_1,x_2)$.

$\begin{array}{|rcll|} \hline x^3 + 5x^2 + px + q &=& 0,\ \text{ the roots are }~ x_1,x_3,x_4. \\ x^3 + 7x^2 + px + r &=& 0,\ \text{ the roots are }~ x_2,x_3,x_4. \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \\ 5+x_1 &=& -(x_3+x_4) \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \\ 7+x_2 &=& -(x_3+x_4) \\\\ 5+x_1 &=& 7+x_2 \\ \mathbf{x_1-x_2} &=& \mathbf{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline x_1x_3+x_1x_4+x_3x_4 = p &=& x_2x_3+x_2x_4+x_3x_4 \\ x_1x_3+x_1x_4 &=& x_2x_3+x_2x_4 \\ x_1(x_3+x_4)-x_2(x_3+x_4) &=& 0 \\ (x_3+x_4)(x_1-x_2) &=& 0 \quad | \quad \mathbf{x_1-x_2 = 2} \\ (x_3+x_4)*2 &=& 0 \quad | \quad : 2 \\ x_3+x_4 &=& 0 \\ \mathbf{x_3} &=& \mathbf{-x_4} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 5 &=& -(x_1-x_4+x_4) \\ 5 &=& -x_1 \\ \mathbf{x_1} &=& \mathbf{-5} \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 7 &=& -(x_2-x_4+x_4) \\ 7 &=& -x_2 \\ \mathbf{x_2} &=& \mathbf{-7} \\ \hline \end{array}$

$(x_1,~x_2) = (-5,~-7)$