Find roots of polynomial $$ p(x) = -y^2+\frac{1}{2} $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= \frac{\sqrt{ 2 }}{ 2 }\\[1 em]y_2 &= - \frac{\sqrt{ 2 }}{ 2 } \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 2 } $.
$$ \begin{aligned} -y^2+\frac{1}{2} & = 0 ~~~ / \cdot \color{blue}{ 2 } \\[1 em] -2y^2+1 & = 0 \end{aligned} $$Step 2:
The solutions of $ -2y^2+1 = 0 $ are: $ y = - \dfrac{\sqrt{ 2 }}{ 2 } ~ \text{and} ~ y = \dfrac{\sqrt{ 2 }}{ 2 }$.
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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