First we need to factor trinomial $ \color{blue}{ x^2-4x+4 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-4x+4 = 0} $.

Step 1:

Both the first and third terms are perfect squares.

$$ x^2 = \left( \color{blue}{ x } \right)^2 ~~ \text{and} ~~ 4 = \left( \color{red}{ 2 } \right)^2 $$

The middle term ( $ -4x $ ) is two times the product of the terms that are squared.

$$ -4x = - 2 \cdot \color{blue}{x} \cdot \color{red}{2} $$

We can conclude that the polynomial $ x^{2}-4x+4 $ is a perfect square trinomial, so we will use the formula below.

$$ A^2 - 2AB + B^2 = (A - B)^2 $$

In this example we have $ \color{blue}{ A = x } $ and $ \color{red}{ B = 2 } $ so,

$$ x^{2}-4x+4 = ( \color{blue}{ x } - \color{red}{ 2 } )^2 $$

Step 2: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} x-2 &= 0 \\ x &= 2 \end{aligned} & ~ & \begin{aligned} x-2 &= 0 \\ x &= 2 \end{aligned} \end{array} $$

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