Solve the Absolute Value Inequality for x |x^2+x-1|=1

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

Step 1

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

$x^{2} + x - 1 = \pm 1$

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.

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $- 1$ .

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

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 $1$ .

$- 1, 2$

Write the factored form using these integers.

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

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.

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

Add $1$ to both sides of the equation.

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

Combine the opposite terms in $x^{2} + x - 1 + 1$ .

Add $- 1$ and $1$ .

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

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

$x^{2} + x = 0$

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

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

$x \cdot x + x = 0$

Raise $x$ to the power of $1$ .

$x \cdot x + x = 0$

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

$x \cdot x + x \cdot 1 = 0$

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

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

$x = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

$x = 0, - 1$

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

$x = 1, - 2, 0, - 1$

Solve the Absolute Value Inequality for x |x^2+x-1|=1

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