Find roots of polynomial $$ p(x) = x^2+9 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3i\\[1 em]x_2 &= -3i \end{aligned} $$Explanation
The solutions of $ x^2+9 = 0 $ are: $ x = 3 i ~ \text{and} ~ x = -3 i $.
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