# How do you find the complex roots of a^2-81=0?

Nov 27, 2016

This quadratic has only two Real roots.

#### Explanation:

This can be factored as a difference of squares, or ${a}^{2} - {b}^{2} = 0$.

${a}^{2} - 81 = 0$

${a}^{2} - {9}^{2} = 0$

$\left(a + 9\right) \left(a - 9\right) = 0$

So the two solutions, both of which are Real, are $a = \pm 9$.