Question
Find roots of polynomial $$ p(x) = z^2+\frac{1}{81} $$
Answer
The roots of polynomial $ p(z) $ are:
$$ \begin{aligned}z_1 &= \dfrac{ 1 }{ 9 }i\\[1 em]z_2 &= -\dfrac{ 1 }{ 9 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 81 } $.
$$ \begin{aligned} z^2+\frac{1}{81} & = 0 ~~~ / \cdot \color{blue}{ 81 } \\[1 em] 81z^2+1 & = 0 \end{aligned} $$Step 2:
The solutions of $ 81z^2+1 = 0 $ are: $ z = \dfrac{ 1 }{ 9 } i ~ \text{and} ~ z = -\dfrac{ 1 }{ 9 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
This page was created using
Polynomial Roots Calculator