Precalculus Examples

$| x^{2} - 2 x | = 8$

Step 1

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

$x^{2} - 2 x = \pm 8$

Step 2

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

Tap for more steps...

Step 2.1

First, use the positive value of the $\pm$ to find the first solution.

$x^{2} - 2 x = 8$

Step 2.2

Subtract $8$ from both sides of the equation.

$x^{2} - 2 x - 8 = 0$

Step 2.3

Factor $x^{2} - 2 x - 8$ using the AC method.

Tap for more steps...

Step 2.3.1

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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Step 2.3.2

Write the factored form using these integers.

$(x - 4) (x + 2) = 0$

Step 2.4

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

$x - 4 = 0$

$x + 2 = 0$

Step 2.5

Set $x - 4$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.6

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

Tap for more steps...

Step 2.6.1

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

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

The final solution is all the values that make $(x - 4) (x + 2) = 0$ true.

$x = 4, - 2$

Step 2.8

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

$x^{2} - 2 x = - 8$

Step 2.9

Add $8$ to both sides of the equation.

$x^{2} - 2 x + 8 = 0$

Step 2.10

Use the quadratic formula to find the solutions.

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

Step 2.11

Substitute the values $a = 1$ , $b = - 2$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 8)}}{2 \cdot 1}$

Step 2.12

Simplify.

Tap for more steps...

Step 2.12.1

Simplify the numerator.

Tap for more steps...

Step 2.12.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Step 2.12.1.2

Multiply $- 4 \cdot 1 \cdot 8$ .

Tap for more steps...

Step 2.12.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 8}}{2 \cdot 1}$

Step 2.12.1.2.2

Multiply $- 4$ by $8$ .

$x = \frac{2 \pm \sqrt{4 - 32}}{2 \cdot 1}$

Step 2.12.1.3

Subtract $32$ from $4$ .

$x = \frac{2 \pm \sqrt{- 28}}{2 \cdot 1}$

Step 2.12.1.4

Rewrite $- 28$ as $- 1 (28)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 28}}{2 \cdot 1}$

Step 2.12.1.5

Rewrite $\sqrt{- 1 (28)}$ as $\sqrt{- 1} \cdot \sqrt{28}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{28}}{2 \cdot 1}$

Step 2.12.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{28}}{2 \cdot 1}$

Step 2.12.1.7

Rewrite $28$ as $2^{2} \cdot 7$ .

Tap for more steps...

Step 2.12.1.7.1

Factor $4$ out of $28$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (7)}}{2 \cdot 1}$

Step 2.12.1.7.2

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Step 2.12.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{7})}{2 \cdot 1}$

Step 2.12.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{7}}{2 \cdot 1}$

Step 2.12.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{7}}{2}$

Step 2.12.3

Simplify $\frac{2 \pm 2 i \sqrt{7}}{2}$ .

$x = 1 \pm i \sqrt{7}$

Step 2.13

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{7}, 1 - i \sqrt{7}$

Step 2.14

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

$x = 4, - 2, 1 + i \sqrt{7}, 1 - i \sqrt{7}$