\(4 x^{2}-4 x+1=(2 x-1)(2 x-1)\)
यदि \(4 x^{2}-4 x+1=0\) तो \(x=1 / 2\) या \(1 / 2\)
\(4 x^{2}-4 x+1\) के शून्यक \( \frac{1}{2}, \frac{1}{2}\)
\(\alpha+\beta=\frac{4}{4}=1\) \(\text {L.H.S. }=\frac{1}{2}+\frac{1}{2}=1\)
\(\alpha \beta=\frac{1}{4} \) \(\text {L.H.S. }=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}\)