Solve for x |x^2+3x-2|=2

$| x^{2} + 3 x - 2 | = 2$

Step 1

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

$x^{2} + 3 x - 2 = \pm 2$

Step 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.

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

Subtract $2$ from both sides of the equation.

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

Subtract $2$ from $- 2$ .

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

Factor $x^{2} + 3 x - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$(x - 1) (x + 4) = 0$

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

$x - 1 = 0$

$x + 4 = 0$

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 $x + 4$ 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 $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

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

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

Add $2$ to both sides of the equation.

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

Combine the opposite terms in $x^{2} + 3 x - 2 + 2$ .

Add $- 2$ and $2$ .

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

Add $x^{2} + 3 x$ and $0$ .

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

Factor $x$ out of $x^{2} + 3 x$ .

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

$x \cdot x + 3 x = 0$

Factor $x$ out of $3 x$ .

$x \cdot x + x \cdot 3 = 0$

Factor $x$ out of $x \cdot x + x \cdot 3$ .

$x (x + 3) = 0$

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

$x = 0$

$x + 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 0, - 3$

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

$x = 1, - 4, 0, - 3$

Solve for x |x^2+3x-2|=2

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