Résoudre une équation du second degré à coefficients réels dans C - TS - Exercice Mathématiques - Kartable

\dfrac{1}{2}z^2+5z+13=0

S=\left\{ -5+i ;-5-i \right\}

S=\left\{ 5+i ;5-i \right\}

S=\left\{ -5+i ;-5+i \right\}

S=\left\{ 5+i ;-5-i \right\}

z^2-2z+4=0

S=\left\{ 1-i\sqrt3{,}1+i\sqrt3 \right\}

S=\left\{ -1-i\sqrt3,-1+i\sqrt3 \right\}

S=\left\{ \dfrac{1}{2}-i\dfrac{\sqrt3}{2}, \dfrac{1}{2}+i\dfrac{\sqrt3}{2} \right\}

S=\left\{ -\dfrac{1}{2}-i\dfrac{\sqrt3}{2}, -\dfrac{1}{2}+i\dfrac{\sqrt3}{2} \right\}

3z^2-3z+1=0

S=\left\{ \dfrac{1}{3}+i\dfrac{\sqrt3}{4}, \dfrac{1}{3}-i\dfrac{\sqrt3}{4} \right\}

S=\left\{ -\dfrac{2}{3}+i\dfrac{\sqrt3}{4}, -\dfrac{2}{3}+i\dfrac{\sqrt3}{4} \right\}

S=\left\{ \dfrac{1}{2}+i\dfrac{\sqrt3}{6}, \dfrac{1}{2}-i\dfrac{\sqrt3}{6} \right\}

S=\left\{ \dfrac{5}{2}+i\dfrac{\sqrt3}{2}, \dfrac{5}{2}-i\dfrac{\sqrt3}{2} \right\}

4z^2+5z+3=0

S=\left\{ -\dfrac{3}{7}+i\dfrac{\sqrt{21}}{8}, -\dfrac{3}{7}-i\dfrac{\sqrt{21}}{8} \right\}

S=\left\{ -\dfrac{3}{4}+i\dfrac{5}{8}, -\dfrac{3}{4}-i\dfrac{5}{8} \right\}

S=\left\{ -\dfrac{5}{8}+i\dfrac{\sqrt{23}}{8}, -\dfrac{5}{8}-i\dfrac{\sqrt{23}}{8} \right\}

S=\left\{ -\dfrac{2}{5}+i\dfrac{5}{7}, -\dfrac{2}{5}-i\dfrac{5}{7} \right\}

6z^2-2z+\dfrac{3}{2}=0

S=\left\{ \dfrac{5}{6}+i\dfrac{\sqrt5}{4}, \dfrac{5}{6}-i\dfrac{\sqrt5}{4} \right\}

S=\left\{ \dfrac{2}{7}+i\dfrac{\sqrt3}{4}, \dfrac{2}{7}-i\dfrac{\sqrt3}{4} \right\}

S=\left\{ \dfrac{1}{6}+i\dfrac{\sqrt2}{3}, \dfrac{1}{6}-i\dfrac{\sqrt2}{3} \right\}

S=\left\{ \dfrac{1}{3}+i\dfrac{2}{3}, \dfrac{1}{3}-i\dfrac{2}{3} \right\}

3z^2-4z+\dfrac{5}{2}=0

S=\left\{ \dfrac{5}{4}+i\dfrac{\sqrt14}{7}, \dfrac{5}{4}-i\dfrac{\sqrt14}{7} \right\}

S=\left\{ \dfrac{1}{5}+i\dfrac{\sqrt14}{4}, \dfrac{1}{5}-i\dfrac{\sqrt14}{4} \right\}

S=\left\{ \dfrac{2}{3}+i\dfrac{\sqrt14}{6}, \dfrac{2}{3}-i\dfrac{\sqrt14}{6} \right\}

S=\left\{ -\dfrac{4}{5}+i\dfrac{\sqrt14}{3}, -\dfrac{4}{5}-i\dfrac{\sqrt14}{3} \right\}