Résoudre A*a+b=8quadquadab=3

$$A \cdot a + b = 8 \quad \quad a b = 3$$

Évaluer (solution complexe)

$$Aa+b=8ab\text{ and }8ab=3$$

Calculer b

$$\left\{\begin{matrix}b=-\frac{\sqrt{6}\left(\sqrt{6-A}+\sqrt{6}\right)A}{4\left(A-2\sqrt{6\left(6-A\right)}-12\right)}\text{, }&a=\frac{\sqrt{6}\left(\sqrt{6-A}+\sqrt{6}\right)}{4A}\text{ and }A\neq 0\text{ and }A\leq 6\\b=\frac{\left(\sqrt{6\left(6-A\right)}-6\right)A}{4\left(A+2\sqrt{6\left(6-A\right)}-12\right)}\text{, }&a=-\frac{\sqrt{576-96A}-24}{16A}\text{ and }A\neq 0\text{ and }A\leq 6\\b=3\text{, }&a=\frac{1}{8}\text{ and }A=0\end{matrix}\right,$$

Calculer a

$$a=\frac{3}{8b}$$

$$A=-\frac{8b\left(b-3\right)}{3}\text{ and }b\neq 0$$