Solve for x |x^2-4|=4-2x

$| x^{2} - 4 | = 4 - 2 x$

Reorder $4$ and $- 2 x$ .

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

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x^{2} - 4 = \pm (- 2 x + 4)$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Add $2 x$ to both sides of the equation.

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

Subtract $4$ from both sides of the equation.

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

Subtract $4$ from $- 4$ .

$x^{2} + 2 x - 8 = 0$

Factor $x^{2} + 2 x - 8$ 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 $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$(x - 2) (x + 4) = 0$

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

$x - 2 = 0$

$x + 4 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

Next, use the negative value of the $\pm$ to find the second solution.

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

Simplify $- (- 2 x + 4)$ .

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

$x^{2} - 4 = - (- 2 x) - 1 \cdot 4$

Multiply.

Multiply $- 2$ by $- 1$ .

$x^{2} - 4 = 2 x - 1 \cdot 4$

Multiply $- 1$ by $4$ .

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

Subtract $2 x$ from both sides of the equation.

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

Add $4$ to both sides of the equation.

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

Combine the opposite terms in $x^{2} - 4 - 2 x + 4$ .

Add $- 4$ and $4$ .

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

Add $x^{2} - 2 x$ and $0$ .

$x^{2} - 2 x = 0$

Factor $x$ out of $x^{2} - 2 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 2 x = 0$

Factor $x$ out of $- 2 x$ .

$x \cdot x + x \cdot - 2 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 2$ .

$x (x - 2) = 0$

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

$x = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $x (x - 2) = 0$ true.

$x = 0, 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 4, 0$

Solve for x |x^2-4|=4-2x

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