$$
\begin{aligned}
&4 x^{2}-4 x+1=0 \\
&\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0
\end{aligned}
$$

$$
\begin{aligned}
&\Rightarrow(2 \mathrm{x}-1)^{2}=0 \quad\left[\because \mathrm{a}^{2}-2 \mathrm{ab}+\mathrm{b}^{2}=(\mathrm{a}-\mathrm{b})^{2}\right] \\
&\Rightarrow(2 \mathrm{x}-1)^{2}=0 \\
&\Rightarrow \mathrm{x}=\frac{1}{2} \text { or } \mathrm{x}=\frac{1}{2} \\
&\text { Sum of zeroes }=\frac{1}{2}+\frac{1}{2}=1=\frac{1}{1}=\frac{-(\text { coefficient of } x)}{\text { (coefficient of } \left.x^{2}\right)} \\
&\text { Product of zeroes }=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=\frac{\text { constant term }}{\text { (coef ficient of } \left.x^{2}\right)}
\end{aligned}
$$