Precalculus Examples

$g (s) = \sqrt{\frac{3 x + 4}{x^{2} - 5}}$

Step 1

Set the radicand in $\sqrt{\frac{3 x + 4}{x^{2} - 5}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{3 x + 4}{x^{2} - 5} \geq 0$

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.

$3 x + 4 = 0$

$x^{2} - 5 = 0$

Step 2.2

Subtract $4$ from both sides of the equation.

$3 x = - 4$

Step 2.3

Divide each term in $3 x = - 4$ by $3$ and simplify.

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Step 2.3.1

Divide each term in $3 x = - 4$ by $3$ .

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $3$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{3}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

Step 2.4

Add $5$ to both sides of the equation.

$x^{2} = 5$

Step 2.5

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

$x = \pm \sqrt{5}$

Step 2.6

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

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Step 2.6.1

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

$x = \sqrt{5}$

Step 2.6.2

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

$x = - \sqrt{5}$

Step 2.6.3

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

$x = \sqrt{5}, - \sqrt{5}$

Step 2.7

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

$x = - \frac{4}{3}$

$x = \sqrt{5}, - \sqrt{5}$

Step 2.8

Consolidate the solutions.

$x = - \frac{4}{3}, \sqrt{5}, - \sqrt{5}$

Step 2.9

Find the domain of $\frac{3 x + 4}{x^{2} - 5}$ .

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Step 2.9.1

Set the denominator in $\frac{3 x + 4}{x^{2} - 5}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 = 0$

Step 2.9.2

Solve for $x$ .

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Step 2.9.2.1

Add $5$ to both sides of the equation.

$x^{2} = 5$

Step 2.9.2.2

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

$x = \pm \sqrt{5}$

Step 2.9.2.3

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

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Step 2.9.2.3.1

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

$x = \sqrt{5}$

Step 2.9.2.3.2

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

$x = - \sqrt{5}$

Step 2.9.2.3.3

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

$x = \sqrt{5}, - \sqrt{5}$

Step 2.9.3

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

$(- \infty, - \sqrt{5}) \cup (- \sqrt{5}, \sqrt{5}) \cup (\sqrt{5}, \infty)$

Step 2.10

Use each root to create test intervals.

$x < - \sqrt{5}$

$- \sqrt{5} < x < - \frac{4}{3}$

$- \frac{4}{3} < x < \sqrt{5}$

$x > \sqrt{5}$

Step 2.11

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 2.11.1

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

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Step 2.11.1.1

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

$x = - 5$

Step 2.11.1.2

Replace $x$ with $- 5$ in the original inequality.

$\frac{3 (- 5) + 4}{{(- 5)}^{2} - 5} \geq 0$

Step 2.11.1.3

The left side $- 0.55$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.11.2

Test a value on the interval $- \sqrt{5} < x < - \frac{4}{3}$ to see if it makes the inequality true.

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Step 2.11.2.1

Choose a value on the interval $- \sqrt{5} < x < - \frac{4}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.11.2.2

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

$\frac{3 (- 2) + 4}{{(- 2)}^{2} - 5} \geq 0$

Step 2.11.2.3

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.11.3

Test a value on the interval $- \frac{4}{3} < x < \sqrt{5}$ to see if it makes the inequality true.

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Step 2.11.3.1

Choose a value on the interval $- \frac{4}{3} < x < \sqrt{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.11.3.2

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

$\frac{3 (0) + 4}{{(0)}^{2} - 5} \geq 0$

Step 2.11.3.3

The left side $- 0.8$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.11.4

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

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Step 2.11.4.1

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

$x = 5$

Step 2.11.4.2

Replace $x$ with $5$ in the original inequality.

$\frac{3 (5) + 4}{{(5)}^{2} - 5} \geq 0$

Step 2.11.4.3

The left side $0.95$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.11.5

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

$x < - \sqrt{5}$ False

$- \sqrt{5} < x < - \frac{4}{3}$ True

$- \frac{4}{3} < x < \sqrt{5}$ False

$x > \sqrt{5}$ True

$x < - \sqrt{5}$ False

$- \sqrt{5} < x < - \frac{4}{3}$ True

$- \frac{4}{3} < x < \sqrt{5}$ False

$x > \sqrt{5}$ True

Step 2.12

The solution consists of all of the true intervals.

$- \sqrt{5} < x \leq - \frac{4}{3}$ or $x > \sqrt{5}$

Step 3

Set the denominator in $\frac{3 x + 4}{x^{2} - 5}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $5$ to both sides of the equation.

$x^{2} = 5$

Step 4.2

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

$x = \pm \sqrt{5}$

Step 4.3

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

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Step 4.3.1

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

$x = \sqrt{5}$

Step 4.3.2

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

$x = - \sqrt{5}$

Step 4.3.3

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

$x = \sqrt{5}, - \sqrt{5}$

Step 5

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

Interval Notation:

$(- \sqrt{5}, - \frac{4}{3}] \cup (\sqrt{5}, \infty)$

Set-Builder Notation:

${x | - \sqrt{5} < x \leq - \frac{4}{3}, x > \sqrt{5}}$

Step 6