Precalculus Examples

$\frac{x}{x^{2} + 2 x - 6} \leq 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 2$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 4.1.3

Add $4$ and $24$ .

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

Step 4.1.4

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

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Step 4.1.4.1

Factor $4$ out of $28$ .

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

Step 4.1.4.2

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

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

Step 4.1.5

Pull terms out from under the radical.

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

Step 4.2

Multiply $2$ by $1$ .

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

Step 4.3

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

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

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 5.1.3

Add $4$ and $24$ .

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

Step 5.1.4

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

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Step 5.1.4.1

Factor $4$ out of $28$ .

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

Step 5.1.4.2

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

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

Step 5.1.5

Pull terms out from under the radical.

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

Step 5.2

Multiply $2$ by $1$ .

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

Step 5.3

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

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

Step 5.4

Change the $\pm$ to $+$ .

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

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 6.1.3

Add $4$ and $24$ .

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

Step 6.1.4

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

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Step 6.1.4.1

Factor $4$ out of $28$ .

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

Step 6.1.4.2

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

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

Step 6.1.5

Pull terms out from under the radical.

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

Step 6.2

Multiply $2$ by $1$ .

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

Step 6.3

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

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

Step 6.4

Change the $\pm$ to $-$ .

$x = - 1 - \sqrt{7}$

Step 7

The final answer is the combination of both solutions.

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

Step 8

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

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

Step 9

Consolidate the solutions.

$x = 0, - 1 + \sqrt{7}, - 1 - \sqrt{7}$

Step 10

Find the domain of $\frac{x}{x^{2} + 2 x - 6}$ .

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Step 10.1

Set the denominator in $\frac{x}{x^{2} + 2 x - 6}$ equal to $0$ to find where the expression is undefined.

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

Step 10.2

Solve for $x$ .

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Step 10.2.1

Use the quadratic formula to find the solutions.

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

Step 10.2.2

Substitute the values $a = 1$ , $b = 2$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 10.2.3

Simplify.

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Step 10.2.3.1

Simplify the numerator.

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Step 10.2.3.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 10.2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 10.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 10.2.3.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 10.2.3.1.3

Add $4$ and $24$ .

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

Step 10.2.3.1.4

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

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Step 10.2.3.1.4.1

Factor $4$ out of $28$ .

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

Step 10.2.3.1.4.2

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

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

Step 10.2.3.1.5

Pull terms out from under the radical.

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

Step 10.2.3.2

Multiply $2$ by $1$ .

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

Step 10.2.3.3

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

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

Step 10.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 10.2.4.1

Simplify the numerator.

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Step 10.2.4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 10.2.4.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 10.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 10.2.4.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 10.2.4.1.3

Add $4$ and $24$ .

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

Step 10.2.4.1.4

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

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Step 10.2.4.1.4.1

Factor $4$ out of $28$ .

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

Step 10.2.4.1.4.2

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

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

Step 10.2.4.1.5

Pull terms out from under the radical.

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

Step 10.2.4.2

Multiply $2$ by $1$ .

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

Step 10.2.4.3

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

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

Step 10.2.4.4

Change the $\pm$ to $+$ .

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

Step 10.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 10.2.5.1

Simplify the numerator.

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Step 10.2.5.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 10.2.5.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 10.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 6}}{2 \cdot 1}$

Step 10.2.5.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 1}$

Step 10.2.5.1.3

Add $4$ and $24$ .

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

Step 10.2.5.1.4

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

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Step 10.2.5.1.4.1

Factor $4$ out of $28$ .

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

Step 10.2.5.1.4.2

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

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

Step 10.2.5.1.5

Pull terms out from under the radical.

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

Step 10.2.5.2

Multiply $2$ by $1$ .

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

Step 10.2.5.3

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

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

Step 10.2.5.4

Change the $\pm$ to $-$ .

$x = - 1 - \sqrt{7}$

Step 10.2.6

The final answer is the combination of both solutions.

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

Step 10.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 1 - \sqrt{7}) \cup (- 1 - \sqrt{7}, - 1 + \sqrt{7}) \cup (- 1 + \sqrt{7}, \infty)$

Step 11

Use each root to create test intervals.

$x < - 1 - \sqrt{7}$

$- 1 - \sqrt{7} < x < 0$

$0 < x < - 1 + \sqrt{7}$

$x > - 1 + \sqrt{7}$

Step 12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 12.1

Test a value on the interval $x < - 1 - \sqrt{7}$ to see if it makes the inequality true.

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Step 12.1.1

Choose a value on the interval $x < - 1 - \sqrt{7}$ and see if this value makes the original inequality true.

$x = - 6$

Step 12.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{- 6}{{(- 6)}^{2} + 2 (- 6) - 6} \leq 0$

Step 12.1.3

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $- 1 - \sqrt{7} < x < 0$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $- 1 - \sqrt{7} < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 12.2.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{- 2}{{(- 2)}^{2} + 2 (- 2) - 6} \leq 0$

Step 12.2.3

The left side $0.\overline{3}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 12.3

Test a value on the interval $0 < x < - 1 + \sqrt{7}$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $0 < x < - 1 + \sqrt{7}$ and see if this value makes the original inequality true.

$x = 1$

Step 12.3.2

Replace $x$ with $1$ in the original inequality.

$\frac{1}{{(1)}^{2} + 2 (1) - 6} \leq 0$

Step 12.3.3

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.4

Test a value on the interval $x > - 1 + \sqrt{7}$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $x > - 1 + \sqrt{7}$ and see if this value makes the original inequality true.

$x = 4$

Step 12.4.2

Replace $x$ with $4$ in the original inequality.

$\frac{4}{{(4)}^{2} + 2 (4) - 6} \leq 0$

Step 12.4.3

The left side $0.\overline{2}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 12.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 1 - \sqrt{7}$ True

$- 1 - \sqrt{7} < x < 0$ False

$0 < x < - 1 + \sqrt{7}$ True

$x > - 1 + \sqrt{7}$ False

$x < - 1 - \sqrt{7}$ True

$- 1 - \sqrt{7} < x < 0$ False

$0 < x < - 1 + \sqrt{7}$ True

$x > - 1 + \sqrt{7}$ False

Step 13

The solution consists of all of the true intervals.

$x < - 1 - \sqrt{7}$ or $0 \leq x < - 1 + \sqrt{7}$

Step 14

Convert the inequality to interval notation.

$(- \infty, - 1 - \sqrt{7}) \cup [0, - 1 + \sqrt{7})$

Step 15