Precalculus Examples

$\frac{(x^{2} + 1) (x - 5)}{x^{2} - 25} \geq 0$

Step 1

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

$x^{2} + 1 = 0$

$x - 5 = 0$

$x^{2} - 25 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 4

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

$x = \pm i$

Step 5

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

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Step 5.1

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

$x = i$

Step 5.2

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

$x = - i$

Step 5.3

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

$x = i, - i$

Step 6

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 8

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 9

Simplify $\pm \sqrt{25}$ .

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Step 9.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 9.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 10

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

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Step 10.1

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

$x = 5$

Step 10.2

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

$x = - 5$

Step 10.3

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

$x = 5, - 5$

Step 11

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

$x = i, - i$

$x = 5$

$x = 5, - 5$

Step 12

Consolidate the solutions.

$x = 5, 5, - 5$

Step 13

Find the domain of $\frac{(x^{2} + 1) (x - 5)}{x^{2} - 25}$ .

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Step 13.1

Set the denominator in $\frac{(x^{2} + 1) (x - 5)}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 13.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 13.2.3

Simplify $\pm \sqrt{25}$ .

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Step 13.2.3.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 13.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 13.2.4

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

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Step 13.2.4.1

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

$x = 5$

Step 13.2.4.2

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

$x = - 5$

Step 13.2.4.3

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

$x = 5, - 5$

Step 13.3

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

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Step 14

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 5$

$x > 5$

Step 15

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 15.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 15.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 15.1.2

Replace $x$ with $- 8$ in the original inequality.

$\frac{({(- 8)}^{2} + 1) ((- 8) - 5)}{{(- 8)}^{2} - 25} \geq 0$

Step 15.1.3

The left side $- 21.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 15.2

Test a value on the interval $- 5 < x < 5$ to see if it makes the inequality true.

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Step 15.2.1

Choose a value on the interval $- 5 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 15.2.2

Replace $x$ with $0$ in the original inequality.

$\frac{({(0)}^{2} + 1) ((0) - 5)}{{(0)}^{2} - 25} \geq 0$

Step 15.2.3

The left side $0.2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 15.3

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 15.3.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 15.3.2

Replace $x$ with $8$ in the original inequality.

$\frac{({(8)}^{2} + 1) ((8) - 5)}{{(8)}^{2} - 25} \geq 0$

Step 15.3.3

The left side $5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 15.4

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

$x < - 5$ False

$- 5 < x < 5$ True

$x > 5$ True

$x < - 5$ False

$- 5 < x < 5$ True

$x > 5$ True

Step 16

The solution consists of all of the true intervals.

$- 5 < x < 5$ or $x > 5$

Step 17

Convert the inequality to interval notation.

$(- 5, 5) \cup (5, \infty)$

Step 18