Find roots of polynomial $$ p(x) = m^2-2m+\frac{225}{100} $$
Answer
The roots of polynomial $ p(m) $ are:
$$ \begin{aligned}m_1 &= 1+\frac{\sqrt{ 5 }}{ 2 }i\\[1 em]m_2 &= 1- \frac{\sqrt{ 5 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} m^2-2m+\frac{225}{100} & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 100m^2-200m+225 & = 0 \end{aligned} $$Step 2:
The solutions of $ 100m^2-200m+225 = 0 $ are: $ m = 1+\dfrac{\sqrt{ 5 }}{ 2 }i ~ \text{and} ~ m = 1-\dfrac{\sqrt{ 5 }}{ 2 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
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