Questions: Résoudre dans C^2 les systèmes suivants : 1) x + y = sqrt(3) xy = 1 2) x + y = 2 xy = 8

Solution Steps

For the first system of equations: \[ \begin{cases}
x + y = \sqrt{3} \\
xy = 1
\end{cases} \] We can express this as a quadratic equation: \[ t^2 - (\sqrt{3})t + 1 = 0 \]

For the second system of equations: \[ \begin{cases}
x + y = 2 \\
xy = 8
\end{cases} \] We can express this as a quadratic equation: \[ t^2 - 2t + 8 = 0 \]

The first quadratic equation is: \[ t^2 - \sqrt{3}t + 1 \] This equation does not factor nicely over the integers, so we will use the quadratic formula to find the roots.

The second quadratic equation is: \[ t^2 - 2t + 8 \] This equation also does not factor nicely over the integers, so we will use the quadratic formula to find the roots.

For the first equation \(t^2 - \sqrt{3}t + 1 = 0\), we apply the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -\sqrt{3}\), and \(c = 1\): \[ t = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3} \pm i}{2} \]

For the second equation \(t^2 - 2t + 8 = 0\), we apply the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -2\), and \(c = 8\): \[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 32}}{2} = \frac{2 \pm \sqrt{-28}}{2} = \frac{2 \pm 2i\sqrt{7}}{2} = 1 \pm i\sqrt{7} \]

The solutions for the first system of equations are: \[ \left( \frac{\sqrt{3} + i}{2}, \frac{\sqrt{3} - i}{2} \right) \]

The solutions for the second system of equations are: \[ (1 + i\sqrt{7}, 1 - i\sqrt{7}) \]

Final Answer