Precalculus Examples

$\sqrt{\frac{x}{x^{2} - 49}}$

Step 1

Set the radicand in $\sqrt{\frac{x}{x^{2} - 49}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{x}{x^{2} - 49} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x = 0$

$x^{2} - 49 = 0$

Step 2.2

Add $49$ to both sides of the equation.

$x^{2} = 49$

Step 2.3

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

$x = \pm \sqrt{49}$

Step 2.4

Simplify $\pm \sqrt{49}$ .

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Step 2.4.1

Rewrite $49$ as $7^{2}$ .

$x = \pm \sqrt{7^{2}}$

Step 2.4.2

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

$x = \pm 7$

Step 2.5

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

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Step 2.5.1

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

$x = 7$

Step 2.5.2

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

$x = - 7$

Step 2.5.3

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

$x = 7, - 7$

Step 2.6

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

$x = 0$

$x = 7, - 7$

Step 2.7

Consolidate the solutions.

$x = 0, 7, - 7$

Step 2.8

Find the domain of $\frac{x}{x^{2} - 49}$ .

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Step 2.8.1

Set the denominator in $\frac{x}{x^{2} - 49}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 49 = 0$

Step 2.8.2

Solve for $x$ .

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Step 2.8.2.1

Add $49$ to both sides of the equation.

$x^{2} = 49$

Step 2.8.2.2

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

$x = \pm \sqrt{49}$

Step 2.8.2.3

Simplify $\pm \sqrt{49}$ .

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Step 2.8.2.3.1

Rewrite $49$ as $7^{2}$ .

$x = \pm \sqrt{7^{2}}$

Step 2.8.2.3.2

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

$x = \pm 7$

Step 2.8.2.4

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

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Step 2.8.2.4.1

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

$x = 7$

Step 2.8.2.4.2

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

$x = - 7$

Step 2.8.2.4.3

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

$x = 7, - 7$

Step 2.8.3

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

$(- \infty, - 7) \cup (- 7, 7) \cup (7, \infty)$

Step 2.9

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 0$

$0 < x < 7$

$x > 7$

Step 2.10

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 2.10.1

Test a value on the interval $x < - 7$ to see if it makes the inequality true.

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Step 2.10.1.1

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

$x = - 10$

Step 2.10.1.2

Replace $x$ with $- 10$ in the original inequality.

$\frac{- 10}{{(- 10)}^{2} - 49} \geq 0$

Step 2.10.1.3

The left side $- 0.19607843$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.10.2

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

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Step 2.10.2.1

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

$x = - 4$

Step 2.10.2.2

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

$\frac{- 4}{{(- 4)}^{2} - 49} \geq 0$

Step 2.10.2.3

The left side $0.\overline{12}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.3

Test a value on the interval $0 < x < 7$ to see if it makes the inequality true.

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Step 2.10.3.1

Choose a value on the interval $0 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 2.10.3.2

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

$\frac{4}{{(4)}^{2} - 49} \geq 0$

Step 2.10.3.3

The left side $- 0.\overline{12}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.10.4

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 2.10.4.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 2.10.4.2

Replace $x$ with $10$ in the original inequality.

$\frac{10}{{(10)}^{2} - 49} \geq 0$

Step 2.10.4.3

The left side $0.19607843$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.5

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

$x < - 7$ False

$- 7 < x < 0$ True

$0 < x < 7$ False

$x > 7$ True

$x < - 7$ False

$- 7 < x < 0$ True

$0 < x < 7$ False

$x > 7$ True

Step 2.11

The solution consists of all of the true intervals.

$- 7 < x \leq 0$ or $x > 7$

Step 3

Set the denominator in $\frac{x}{x^{2} - 49}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 49 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $49$ to both sides of the equation.

$x^{2} = 49$

Step 4.2

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

$x = \pm \sqrt{49}$

Step 4.3

Simplify $\pm \sqrt{49}$ .

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Step 4.3.1

Rewrite $49$ as $7^{2}$ .

$x = \pm \sqrt{7^{2}}$

Step 4.3.2

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

$x = \pm 7$

Step 4.4

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

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Step 4.4.1

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

$x = 7$

Step 4.4.2

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

$x = - 7$

Step 4.4.3

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

$x = 7, - 7$

Step 5

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

Interval Notation:

$(- 7, 0] \cup (7, \infty)$

Set-Builder Notation:

${x | - 7 < x \leq 0, x > 7}$

Step 6