Solve for x 1=x^2-4x+4

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

Rewrite the equation as $x^{2} - 4 x + 4 = 1$ .

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $4$ .

$x^{2} - 4 x + 3 = 0$

Factor $x^{2} - 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$(x - 3) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Solve for x 1=x^2-4x+4

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