Question
Solve the following quadratic equation by factorization method $2 x^2-x+\frac{1}{8}=0$
✓
Answer
$\begin{aligned} & 2 x^2-x+\frac{1}{8}=0 \\ & 16 x^2-8 x+1=0 \quad \ldots(\text { multiply by } 8)\end{aligned}$
$
\begin{aligned}
& 16 x^2-4 x-4 x+1 \\
& 4 x(4 x-1)-1(4 x-1)=0 \\
& (4 x-1)(4 x-1)=0 \\
& 4 x=1,4 x=1 \\
& x=\frac{1}{4}, x=\frac{1}{4}
\end{aligned}
$
The roots are $\frac{1}{4}$ and $\frac{1}{4}$
$
\begin{aligned}
& 16 x^2-4 x-4 x+1 \\
& 4 x(4 x-1)-1(4 x-1)=0 \\
& (4 x-1)(4 x-1)=0 \\
& 4 x=1,4 x=1 \\
& x=\frac{1}{4}, x=\frac{1}{4}
\end{aligned}
$
The roots are $\frac{1}{4}$ and $\frac{1}{4}$
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