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

Question

1975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T05:28:52

This quadratic has only two Real roots.

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

$a^2-81=0$

$a^2-9^2=0$

$(a+9)(a-9)=0$

So the two solutions, both of which are Real, are $a=+-9$ .

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