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

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Answer 1 · 2016-11-27T05:28:52

This quadratic has only two Real roots.

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

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

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

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

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

How do you find the complex roots of a2−81=0a^2-81=0?