Question

Solve each of the following quadratic equations:

$2x^2-x+\frac{1}{8}=0$

Answer 1

We write, $-x=-\frac{x}{2}-\frac{x}{2}$ as $2x^2\times \frac{1}{8}=\frac{x^2}{4}=\left(-\frac{x}{2}\right)\times \left(-\frac{x}{2}\right)$
$$\begin{gathered} \therefore 2x^2-x+\frac{1}{8}=0 \\ \Rightarrow 2x^2-\frac{x}{2}-\frac{x}{2}+\frac{1}{8}=0 \\ \Rightarrow 2x\left(x-\frac{1}{4}\right)-\frac{1}{2}\left(x-\frac{1}{4}\right)=0 \\ \Rightarrow \left(x-\frac{1}{4}\right)\left(2x-\frac{1}{2}\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-\frac{1}{4}=0\mathrm{or}2x-\frac{1}{2}=0 \\ \Rightarrow x=\frac{1}{4}\mathrm{or}x=\frac{1}{4} \end{gathered}$$
Hence, $\frac{1}{4}$ is the repeated root of the given equation.