Solve by Factoring 4x^2-4x+1=0

$4 x^{2} - 4 x + 1 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

${(2 x)}^{2} - 4 x + 1 = 0$

Rewrite $1$ as $1^{2}$ .

${(2 x)}^{2} - 4 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

${(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

${(2 x - 1)}^{2} = 0$

Step 2

Set the $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 3

Solve for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$