Now the The Quadratic Formula never fails. When the discriminant $b^2 - 4ac$ is negative, its square root is imaginary, and the formula hands you two complex solutions. You have the tools from Adding and Multiplying Complex Numbers to simplify them.
Idea clave
If $b^2 - 4ac < 0$, the solutions of $ax^2 + bx + c = 0$ are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $\sqrt{b^2 - 4ac} = \sqrt{4ac - b^2}\,i$. They form a conjugate pair $p + qi$ and $p - qi$.
Por qué funciona
The formula comes from completing the square, which never needed the discriminant to be positive. When it is negative, the $\pm$ adds and subtracts the same imaginary amount from the same real number $-\frac{b}{2a}$, so the two answers are conjugates. On the graph this means the parabola never touches the $x$-axis: there are no real zeros, but the equation still has two solutions off the real line.
Worked examples
Ejemplo resuelto: A conjugate pair
Solve $x^2 - 2x + 5 = 0$.
- $a = 1$, $b = -2$, $c = 5$. Discriminant: $4 - 20 = -16$.
- $\sqrt{-16} = 4i$.
- $x = \frac{2 \pm 4i}{2} = 1 \pm 2i$.
Respuesta: $1 + 2i$ and $1 - 2i$
Ejemplo resuelto: Check one solution
Show that $x = 1 + 2i$ satisfies $x^2 - 2x + 5 = 0$.
- $(1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i$.
- $-2(1 + 2i) = -2 - 4i$.
- Add: $-3 + 4i - 2 - 4i + 5 = 0$.
Respuesta: It checks.
Ejemplo resuelto: A root that does not simplify
Solve $2x^2 + 2x + 1 = 0$.
- Discriminant: $4 - 8 = -4$, and $\sqrt{-4} = 2i$.
- $x = \frac{-2 \pm 2i}{4} = -\frac{1}{2} \pm \frac{1}{2}i$.
Respuesta: $-\frac{1}{2} + \frac{1}{2}i$ and $-\frac{1}{2} - \frac{1}{2}i$
Error común
Stopping at "no solution." That answer is right only if the question asks for real solutions. Another slip is dividing only one term by $2a$: in $\frac{2 \pm 4i}{2}$ both the 2 and the $4i$ get divided, giving $1 \pm 2i$, not $1 \pm 4i$.