## Solving quadratic equations with real coefficients

Any quadratic equation of the form $az^2 + bz + c = 0$, where $a,b,c \in \mathbb{R}$ has solutions $\lambda$ and $\mu$, which satisfy one of the following conditions

- $\lambda$ and $\mu$ are distinct real numbers, in this case, $b^2 - 4ac \gt 0$
- $\lambda = \mu$ and $\lambda \in \mathbb{R}$, in this case, $b^2 - 4ac= 0$
- $\lambda = \overline{\mu}$, in this case, $b^2 - 4ac \lt 0$

## Quadratic equations with real coefficients 1

Determine the solutions of the equation $$ z^2 + 2z + 10 = 0 $$

solution - press button to display

$$ \begin{align} z &= \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \\ &= \frac{-2 \pm \sqrt{4-40}}{2} \\ &= \frac{-2 \pm 6i}{2} \\ &= -1 \pm 3i \end{align}$$ the solutions are therefore $z = -1 + 3i$ and $z = -1 - 3i$

## Quadratic equations with real coefficients 2

Determine the solutions of the quadratic equation $$ z^2 + 2\sqrt{2}z + 6 = 0 $$

solution - press button to display

The solutions can be determined using the quadratic formula. $$ \begin{align} z &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-2\sqrt{2} \pm \sqrt{8 - 24}}{2} \\ &= -\sqrt{2} \pm 2i \end{align} $$ The solutions of the equation are therefore $z = -\sqrt{2} + 2i$ and $z = -\sqrt{2} - 2i$