Quadratics by Completing the Square (Complex Roots)

For

$x^2 + px + q = 0$

, completing the square gives:

$\left(x + \frac{p}{2}\right)^2 = \left(\frac{p}{2}\right)^2 - q$

When the right-hand side is negative — say it equals

$-D$

for some positive

$D$

— the roots are complex:

$x = -\frac{p}{2} \pm i\sqrt{D}$

Example:

$x^2 + 2x + 5 = 0$

. Here

$(x+1)^2 = 1-5 = -4$

, so

$x + 1 = \pm 2i$

and

$x = -1 \pm 2i$

.