Precalculus Examples

$h (x) = \sqrt{\frac{x + 2}{x^{2} - 5 x}}$

Step 1

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

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

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.

$x + 2 = 0$

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

Step 2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.3

Factor $x$ out of $x^{2} - 5 x$ .

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Step 2.3.1

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

$x \cdot x - 5 x = 0$

Step 2.3.2

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Step 2.3.3

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

Step 2.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

Step 2.5

Set $x$ equal to $0$ .

$x = 0$

Step 2.6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.7

The final solution is all the values that make $x (x - 5) = 0$ true.

$x = 0, 5$

Step 2.8

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

$x = - 2$

$x = 0, 5$

Step 2.9

Consolidate the solutions.

$x = - 2, 0, 5$

Step 2.10

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

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Step 2.10.1

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

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

Step 2.10.2

Solve for $x$ .

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Step 2.10.2.1

Factor $x$ out of $x^{2} - 5 x$ .

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Step 2.10.2.1.1

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

$x \cdot x - 5 x = 0$

Step 2.10.2.1.2

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Step 2.10.2.1.3

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

Step 2.10.2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

Step 2.10.2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.10.2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.10.2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.10.2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.10.2.5

The final solution is all the values that make $x (x - 5) = 0$ true.

$x = 0, 5$

Step 2.10.3

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

$(- \infty, 0) \cup (0, 5) \cup (5, \infty)$

Step 2.11

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 5$

$x > 5$

Step 2.12

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 2.12.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 2.12.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 2.12.1.2

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

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

Step 2.12.1.3

The left side $- 0.0\overline{5}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.12.2

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

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Step 2.12.2.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 2.12.2.2

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

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

Step 2.12.2.3

The left side $0.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.3

Test a value on the interval $0 < x < 5$ to see if it makes the inequality true.

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Step 2.12.3.1

Choose a value on the interval $0 < x < 5$ and see if this value makes the original inequality true.

$x = 2$

Step 2.12.3.2

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

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

Step 2.12.3.3

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

False

Step 2.12.4

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

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Step 2.12.4.1

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

$x = 8$

Step 2.12.4.2

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

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

Step 2.12.4.3

The left side $0.41\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.5

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

$x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 5$ False

$x > 5$ True

$x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 5$ False

$x > 5$ True

Step 2.13

The solution consists of all of the true intervals.

$- 2 \leq x < 0$ or $x > 5$

Step 3

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

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

Step 4

Solve for $x$ .

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Step 4.1

Factor $x$ out of $x^{2} - 5 x$ .

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Step 4.1.1

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

$x \cdot x - 5 x = 0$

Step 4.1.2

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Step 4.1.3

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

Step 4.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 5 = 0$

Step 4.3

Set $x$ equal to $0$ .

$x = 0$

Step 4.4

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 4.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4.5

The final solution is all the values that make $x (x - 5) = 0$ true.

$x = 0, 5$

Step 5

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

Interval Notation:

$[- 2, 0) \cup (5, \infty)$

Set-Builder Notation:

${x | - 2 \leq x < 0, x > 5}$

Step 6