Precalculus Examples

$| x + 1 | = x^{2} - 5$

Step 1

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

$x + 1 = \pm (x^{2} - 5)$

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.

$x + 1 = x^{2} - 5$

Step 2.2

Subtract $x^{2}$ from both sides of the equation.

$x + 1 - x^{2} = - 5$

Step 2.3

Add $5$ to both sides of the equation.

$x + 1 - x^{2} + 5 = 0$

Step 2.4

Add $1$ and $5$ .

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

Step 2.5

Factor the left side of the equation.

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Step 2.5.1

Factor $- 1$ out of $x - x^{2} + 6$ .

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Step 2.5.1.1

Reorder $x$ and $- x^{2}$ .

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

Step 2.5.1.2

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

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

Step 2.5.1.3

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 6 = 0$

Step 2.5.1.4

Rewrite $6$ as $- 1 (- 6)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 6 = 0$

Step 2.5.1.5

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

$- (x^{2} - x) - 1 \cdot - 6 = 0$

Step 2.5.1.6

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 6)$ .

$- (x^{2} - x - 6) = 0$

Step 2.5.2

Factor.

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Step 2.5.2.1

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

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Step 2.5.2.1.1

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 $- 1$ .

$- 3, 2$

Step 2.5.2.1.2

Write the factored form using these integers.

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

Step 2.5.2.2

Remove unnecessary parentheses.

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

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 $0$ .

$x - 3 = 0$

$x + 2 = 0$

Step 2.7

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

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Step 2.7.1

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

$x - 3 = 0$

Step 2.7.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.8

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

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Step 2.8.1

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

$x + 2 = 0$

Step 2.8.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.9

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

$x = 3, - 2$

Step 2.10

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

$x + 1 = - (x^{2} - 5)$

Step 2.11

Simplify $- (x^{2} - 5)$ .

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Step 2.11.1

Rewrite.

$x + 1 = 0 + 0 - (x^{2} - 5)$

Step 2.11.2

Simplify by adding zeros.

$x + 1 = - (x^{2} - 5)$

Step 2.11.3

Apply the distributive property.

$x + 1 = - x^{2} - - 5$

Step 2.11.4

Multiply $- 1$ by $- 5$ .

$x + 1 = - x^{2} + 5$

Step 2.12

Add $x^{2}$ to both sides of the equation.

$x + 1 + x^{2} = 5$

Step 2.13

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

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Step 2.13.1

Subtract $5$ from both sides of the equation.

$x + 1 + x^{2} - 5 = 0$

Step 2.13.2

Subtract $5$ from $1$ .

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

Step 2.14

Use the quadratic formula to find the solutions.

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

Step 2.15

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

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

Step 2.16

Simplify.

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Step 2.16.1

Simplify the numerator.

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Step 2.16.1.1

One to any power is one.

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

Step 2.16.1.2

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

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Step 2.16.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.16.1.2.2

Multiply $- 4$ by $- 4$ .

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

Step 2.16.1.3

Add $1$ and $16$ .

$x = \frac{- 1 \pm \sqrt{17}}{2 \cdot 1}$

Step 2.16.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{17}}{2}$

Step 2.17

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{17}}{2}, - \frac{1 + \sqrt{17}}{2}$

Step 2.18

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

$x = 3, - 2, - \frac{1 - \sqrt{17}}{2}, - \frac{1 + \sqrt{17}}{2}$

Step 3

Exclude the solutions that do not make $| x + 1 | = x^{2} - 5$ true.

$x = 3, - \frac{1 + \sqrt{17}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = 3, - \frac{1 + \sqrt{17}}{2}$

Decimal Form:

$x = 3, - 2.56155281 \dots$