ដោះស្រាយ {l}{alphabeta^{2}+alpha^2beta=alphabeta(alpha+beta)=(-6)*2=-12}{alpha^2+beta^2=(alpha+beta)^2-2alphabeta}

$$\left. \begin{array} { l } { \alpha \beta ^ { 2 } + \alpha ^ { 2 } \beta = \alpha \beta ( \alpha + \beta ) = ( - 6 ) \times 2 = - 12 } \\ { \alpha ^ { 2 } + \beta ^ { 2 } = ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta } \end{array} \right.$$

វាយតម្លៃ (complex solution)

$$\alpha \beta \left(\alpha +\beta \right)=-12,\ \text{true}$$

ដោះស្រាយសម្រាប់ β

$$\left\{\begin{matrix}\beta =\frac{\sqrt{\alpha ^{2}-\frac{48}{\alpha }}-\alpha }{2}\text{; }\beta =\frac{-\sqrt{\alpha ^{2}-\frac{48}{\alpha }}-\alpha }{2}\text{, }&\alpha \geq 2\sqrt[3]{6}\\\beta =\frac{\sqrt{\alpha ^{4}-48\alpha }}{2\alpha }-\frac{\alpha }{2}\text{; }\beta =-\frac{\sqrt{\alpha ^{4}-48\alpha }}{2\alpha }-\frac{\alpha }{2}\text{, }&\alpha <0\end{matrix}\right.$$

ដោះស្រាយសម្រាប់ α

$$\left\{\begin{matrix}\alpha =\frac{\sqrt{\beta ^{2}-\frac{48}{\beta }}-\beta }{2}\text{; }\alpha =\frac{-\sqrt{\beta ^{2}-\frac{48}{\beta }}-\beta }{2}\text{, }&\beta \geq 2\sqrt[3]{6}\\\alpha =\frac{\sqrt{\beta ^{4}-48\beta }}{2\beta }-\frac{\beta }{2}\text{; }\alpha =-\frac{\sqrt{\beta ^{4}-48\beta }}{2\beta }-\frac{\beta }{2}\text{, }&\beta <0\end{matrix}\right.$$