Precalculus Examples

$\frac{x^{2} - 49}{x - 9} \leq 0$

Step 1

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

$x^{2} - 49 = 0$

$x - 9 = 0$

Step 2

Add $49$ to both sides of the equation.

$x^{2} = 49$

Step 3

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

$x = \pm \sqrt{49}$

Step 4

Simplify $\pm \sqrt{49}$ .

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Step 4.1

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

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

Step 4.2

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

$x = \pm 7$

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.

$x = 7$

Step 5.2

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

$x = - 7$

Step 5.3

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

$x = 7, - 7$

Step 6

Add $9$ to both sides of the equation.

$x = 9$

Step 7

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

$x = 7, - 7$

$x = 9$

Step 8

Consolidate the solutions.

$x = 7, - 7, 9$

Step 9

Find the domain of $\frac{x^{2} - 49}{x - 9}$ .

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Step 9.1

Set the denominator in $\frac{x^{2} - 49}{x - 9}$ equal to $0$ to find where the expression is undefined.

$x - 9 = 0$

Step 9.2

Add $9$ to both sides of the equation.

$x = 9$

Step 9.3

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

$(- \infty, 9) \cup (9, \infty)$

Step 10

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 7$

$7 < x < 9$

$x > 9$

Step 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 11.1

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

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Step 11.1.1

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

$x = - 10$

Step 11.1.2

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

$\frac{{(- 10)}^{2} - 49}{(- 10) - 9} \leq 0$

Step 11.1.3

The left side $- 2.68421052$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 11.2

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

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Step 11.2.1

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

$x = 0$

Step 11.2.2

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

$\frac{{(0)}^{2} - 49}{(0) - 9} \leq 0$

Step 11.2.3

The left side $5.\overline{4}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $7 < x < 9$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $7 < x < 9$ and see if this value makes the original inequality true.

$x = 8$

Step 11.3.2

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

$\frac{{(8)}^{2} - 49}{(8) - 9} \leq 0$

Step 11.3.3

The left side $- 15$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 9$ to see if it makes the inequality true.

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Step 11.4.1

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Step 11.4.2

Replace $x$ with $12$ in the original inequality.

$\frac{{(12)}^{2} - 49}{(12) - 9} \leq 0$

Step 11.4.3

The left side $31.\overline{6}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 7$ True

$- 7 < x < 7$ False

$7 < x < 9$ True

$x > 9$ False

$x < - 7$ True

$- 7 < x < 7$ False

$7 < x < 9$ True

$x > 9$ False

Step 12

The solution consists of all of the true intervals.

$x \leq - 7$ or $7 \leq x < 9$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 7] \cup [7, 9)$

Step 14