Do in completing the square method.
[tex] {4x}^{2} - 4x + 1 = 0 \\ 4x^{2}-4x=-1 \\ \frac{4x^{2}-4x}{4}=\frac{-1}{4} \\ x^{2}+\frac{-4}{4}x=\frac{-1}{4} \\ x^{2}-x=\frac{-1}{4} \\ x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{4}+\left(-\frac{1}{2}\right)^{2} \\ x^{2}-x+\frac{1}{4}=\frac{-1+1}{4} \\ x^{2}-x+\frac{1}{4}=0 \\ \left(x-\frac{1}{2}\right)^{2}=0 \\ \sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{0} [/tex]
Now simplify.
[tex]x-\frac{1}{2}=0 \\ x-\frac{1}{2}=0 \\ \\ \rightarrow \: x=\frac{1}{2} \\ \rightarrow \: x=\frac{1}{2} [/tex]
So, answer is..
[tex] \boxed{x=\frac{1}{2} }[/tex]
Now rewrite the equation.
[tex]4 x ^ { 2 } - 4 x + 1 = 0 \\ \boxed{4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 }[/tex]
Verification :-
[tex]4 x ^ { 2 } - 4 x + 1 = 0 \\ 4\times \left(\frac{1}{4}\right)-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-4\times \left(\frac{1}{2}\right)+1=0 \\ 1-\frac{4}{2}+1=0 \\ 1-2+1=0 \\ -1+1=0 \\ \underline{ 0 = 0} \: \\ \dashrightarrow \text{true}[/tex]
