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

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

Rewrite the equation as $2 - | x | = x^{2}$ .

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

Subtract $2$ from both sides of the equation.

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

Divide each term in $- | x | = x^{2} - 2$ by $- 1$ and simplify.

Divide each term in $- | x | = x^{2} - 2$ by $- 1$ .

$\frac{- | x |}{- 1} = \frac{x^{2}}{- 1} + \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{| x |}{1} = \frac{x^{2}}{- 1} + \frac{- 2}{- 1}$

Divide $| x |$ by $1$ .

$| x | = \frac{x^{2}}{- 1} + \frac{- 2}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$| x | = - 1 \cdot x^{2} + \frac{- 2}{- 1}$

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$| x | = - x^{2} + \frac{- 2}{- 1}$

Divide $- 2$ by $- 1$ .

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

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

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

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.

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

Add $x^{2}$ to both sides of the equation.

$x + x^{2} = 2$

Subtract $2$ from both sides of the equation.

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

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Factor $u + u^{2} - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$(u - 1) (u + 2) = 0$

Replace all occurrences of $u$ with $x$ .

$(x - 1) (x + 2) = 0$

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

$x - 1 = 0$

$x + 2 = 0$

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.

$x = 1$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

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

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

Simplify $- (- x^{2} + 2)$ .

Apply the distributive property.

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

Multiply $- - x^{2}$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x^{2}$ by $1$ .

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

Multiply $- 1$ by $2$ .

$x = x^{2} - 2$

Subtract $x^{2}$ from both sides of the equation.

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

Add $2$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 2$ .

Reorder $x$ and $- x^{2}$ .

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

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $x$ .

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

Rewrite $2$ as $- 1 (- 2)$ .

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

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

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

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 2)$ .

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

Factor.

Factor $x^{2} - x - 2$ 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$- ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 1) = 0$

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

$x - 2 = 0$

$x + 1 = 0$

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 $x + 1$ equal to $0$ and solve for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

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

$x = 1, - 2, 2, - 1$

Exclude the solutions that do not make $x^{2} = 2 - | x |$ true.

$x = 1, - 1$

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

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