4x^2 − 4x + 1

= 4(x^2 − x + 1/4)

= (1 - 2x)^2

= (2x - 1)^2

4x^2 − 4x + 1 = 0

(2x − 1)^2 = 0

4x² − 4x + 1 = 0

4(x² - x) + 1 = 0

4(x² - x + (-1/2)² - (-1/2)²) + 1 = 0

4(x - 1/2)² -4/4 + 1 = 0

4(x - 1/2)² = 0

(x - 1/2)² = 0

Subtract 1 from both sides:

=> 4x^2-4x = -1

Divide everything by 4:

=> x^2-x = -1/4

Complete the square.

The form is a^2-bx = -c

Add (b/2)^2 on both sides (for your question, b = -1):

=> x^2-x+(-1/2)^2 = -1/4+(-1/2)^2

Factor the left side:

=> (x-1/2)^2 = 0

4x^2−4x+1=0. Write it as

4x^2 +4x +1= 8x

(2x+1)²= 8x

(x+1/2)²= 2x. This is ghe answer

4x^2−4x+1=0

(2x)² −2*2x*1+1²=0

i.e. (2x−1)²=0

or 2²{x+(−1/4)}²=0

or {x+(−1/4)}²=0

4x² - 4x + 1 = 0

x² - x + (1/4) = 0

x² - x = - (1/4)

x² - x + (1/2)² = - (1/4) + (1/2)²

x² - x + (1/2)² = - (1/4) + (1/4)

[x - (1/2)]² = 0 → to go further

x - (1/2) = 0

x = 1/2

4[x^2-x]=-1

4[x^2-x+1/4] = -1+1

4(x-1/2)^2 = 0

(x-1/2)^2 = 0

4x^2 - 4x + 1 = 0

(2x - 1)^2 = 0

x = 1/2

4x^2 − 4x + 1 = (2x - 1)^2 = 2(x - 1/2)^2 = 0

So, ( x - 1/2 )^2 = 0

4 [ x^2 - x ] = - 1

x^2 - x = - 1/4

x^2 - x + 1/4 = 0

[ x - 1/2 ]^2 = 0