Solve for x x^2-3x-2=|x-1|

$x^{2} - 3 x - 2 = | x - 1 |$

Rewrite the equation as $| x - 1 | = x^{2} - 3 x - 2$ .

$| x - 1 | = x^{2} - 3 x - 2$

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

$x - 1 = \pm (x^{2} - 3 x - 2)$

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.

$x - 1 = x^{2} - 3 x - 2$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 3 x - 2 = x - 1$

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$x^{2} - 3 x - 2 - x = - 1$

Subtract $x$ from $- 3 x$ .

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

Move all terms to the left side of the equation and simplify.

Add $1$ to both sides of the equation.

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

Add $- 2$ and $1$ .

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{4 \pm \sqrt{16 + 4}}{2 \cdot 1}$

Add $16$ and $4$ .

$x = \frac{4 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{4 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{5}}{2}$ .

$x = 2 \pm \sqrt{5}$

The final answer is the combination of both solutions.

$x = 2 + \sqrt{5}, 2 - \sqrt{5}$

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

$x - 1 = - (x^{2} - 3 x - 2)$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- (x^{2} - 3 x - 2) = x - 1$

Simplify $- (x^{2} - 3 x - 2)$ .

Rewrite.

$0 + 0 - (x^{2} - 3 x - 2) = x - 1$

Simplify by adding zeros.

$- (x^{2} - 3 x - 2) = x - 1$

Apply the distributive property.

$- x^{2} - (- 3 x) - - 2 = x - 1$

Simplify.

Multiply $- 3$ by $- 1$ .

$- x^{2} + 3 x - - 2 = x - 1$

Multiply $- 1$ by $- 2$ .

$- x^{2} + 3 x + 2 = x - 1$

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$- x^{2} + 3 x + 2 - x = - 1$

Subtract $x$ from $3 x$ .

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

Add $1$ to both sides of the equation.

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

Add $2$ and $1$ .

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 2 x + 3$ .

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

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

Factor $- 1$ out of $2 x$ .

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

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

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

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

Factor.

Factor $x^{2} - 2 x - 3$ 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 $- 2$ .

$- 3, 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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 $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 3, - 1$

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

$x = 2 + \sqrt{5}, 2 - \sqrt{5}, 3, - 1$

Exclude the solutions that do not make $x^{2} - 3 x - 2 = | x - 1 |$ true.

$x = 2 + \sqrt{5}, - 1$

The result can be shown in multiple forms.

Exact Form:

$x = 2 + \sqrt{5}, - 1$

Decimal Form:

$x = 4.23606797 \dots, - 1$

Solve for x x^2-3x-2=|x-1|

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