Precalculus Examples

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

Step 1

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

$x^{2} - 4 = \pm (x - 2)$

Step 2

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

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Step 2.1

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

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

Step 2.2

Subtract $x$ from both sides of the equation.

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

Step 2.3

Add $2$ to both sides of the equation.

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

Step 2.4

Add $- 4$ and $2$ .

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

Step 2.5

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

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Step 2.5.1

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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 2.5.2

Write the factored form using these integers.

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

Step 2.6

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

$x - 2 = 0$

$x + 1 = 0$

Step 2.7

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

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Step 2.7.1

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

$x - 2 = 0$

Step 2.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.8

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.8.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.8.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.9

The final solution is all the values that make $(x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Step 2.10

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

$x^{2} - 4 = - (x - 2)$

Step 2.11

Simplify $- (x - 2)$ .

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Step 2.11.1

Rewrite.

$x^{2} - 4 = 0 + 0 - (x - 2)$

Step 2.11.2

Simplify by adding zeros.

$x^{2} - 4 = - (x - 2)$

Step 2.11.3

Apply the distributive property.

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

Step 2.11.4

Multiply $- 1$ by $- 2$ .

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

Step 2.12

Add $x$ to both sides of the equation.

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

Step 2.13

Subtract $2$ from both sides of the equation.

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

Step 2.14

Subtract $2$ from $- 4$ .

$x^{2} + x - 6 = 0$

Step 2.15

Factor $x^{2} + x - 6$ using the AC method.

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Step 2.15.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.15.2

Write the factored form using these integers.

$(x - 2) (x + 3) = 0$

Step 2.16

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

$x - 2 = 0$

$x + 3 = 0$

Step 2.17

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

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Step 2.17.1

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

$x - 2 = 0$

Step 2.17.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.18

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

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Step 2.18.1

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

$x + 3 = 0$

Step 2.18.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.19

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 2.20

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

$x = 2, - 1, - 3$

Step 3

Exclude the solutions that do not make $| x^{2} - 4 | = x - 2$ true.

$x = 2$