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

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

Step 1

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

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

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.

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

Subtract $1$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $4$ and $4$ .

$x = \frac{2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = 1 \pm \sqrt{2}$

The final answer is the combination of both solutions.

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

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

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

Add $1$ to both sides of the equation.

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 2.41421356 \dots, - 0.41421356 \dots, 1$

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

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