Question
Find roots of polynomial $$ p(x) = x^2-4x+\frac{399}{100} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 21 }{ 10 }\\[1 em]x_2 &= \dfrac{ 19 }{ 10 } \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} x^2-4x+\frac{399}{100} & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 100x^2-400x+399 & = 0 \end{aligned} $$Step 2:
The solutions of $ 100x^2-400x+399 = 0 $ are: $ x = \dfrac{ 19 }{ 10 } ~ \text{and} ~ x = \dfrac{ 21 }{ 10 }$.
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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