Precalculus Examples

$\frac{x^{2} (4 + x) (x - 9)}{(x + 2) (x - 7)} \geq 0$

Step 1

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

$x^{2} = 0$

$4 + x = 0$

$x - 9 = 0$

$x + 2 = 0$

$x - 7 = 0$

Step 2

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

$x = \pm \sqrt{0}$

Step 3

Simplify $\pm \sqrt{0}$ .

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Step 3.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 3.2

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

$x = \pm 0$

Step 3.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

Add $9$ to both sides of the equation.

$x = 9$

Step 6

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

Add $7$ to both sides of the equation.

$x = 7$

Step 8

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

$x = 0$

$x = - 4$

$x = 9$

$x = - 2$

$x = 7$

Step 9

Consolidate the solutions.

$x = 0, - 4, 9, - 2, 7$

Step 10

Find the domain of $\frac{x^{2} (4 + x) (x - 9)}{(x + 2) (x - 7)}$ .

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Step 10.1

Set the denominator in $\frac{x^{2} (4 + x) (x - 9)}{(x + 2) (x - 7)}$ equal to $0$ to find where the expression is undefined.

$(x + 2) (x - 7) = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

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

$x + 2 = 0$

$x - 7 = 0$

Step 10.2.2

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 10.2.2.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 10.2.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 10.2.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 10.2.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 10.2.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 10.2.4

The final solution is all the values that make $(x + 2) (x - 7) = 0$ true.

$x = - 2, 7$

Step 10.3

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

$(- \infty, - 2) \cup (- 2, 7) \cup (7, \infty)$

Step 11

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$- 2 < x < 0$

$0 < x < 7$

$7 < x < 9$

$x > 9$

Step 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 12.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

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Step 12.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 12.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{{(- 6)}^{2} (4 - 6) ((- 6) - 9)}{((- 6) + 2) ((- 6) - 7)} \geq 0$

Step 12.1.3

The left side $20.\overline{769230}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

Step 12.2.2

Replace $x$ with $- 3$ in the original inequality.

$\frac{{(- 3)}^{2} (4 - 3) ((- 3) - 9)}{((- 3) + 2) ((- 3) - 7)} \geq 0$

Step 12.2.3

The left side $- 10.8$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.3

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

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Step 12.3.1

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

$x = - 1$

Step 12.3.2

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

$\frac{{(- 1)}^{2} (4 - 1) ((- 1) - 9)}{((- 1) + 2) ((- 1) - 7)} \geq 0$

Step 12.3.3

The left side $3.75$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.4

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

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Step 12.4.1

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

$x = 4$

Step 12.4.2

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

$\frac{{(4)}^{2} (4 + 4) ((4) - 9)}{((4) + 2) ((4) - 7)} \geq 0$

Step 12.4.3

The left side $35.\overline{5}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.5

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

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Step 12.5.1

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

$x = 8$

Step 12.5.2

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

$\frac{{(8)}^{2} (4 + 8) ((8) - 9)}{((8) + 2) ((8) - 7)} \geq 0$

Step 12.5.3

The left side $- 76.8$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.6

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

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Step 12.6.1

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

$x = 12$

Step 12.6.2

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

$\frac{{(12)}^{2} (4 + 12) ((12) - 9)}{((12) + 2) ((12) - 7)} \geq 0$

Step 12.6.3

The left side $98.74285714$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.7

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

$x < - 4$ True

$- 4 < x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 7$ True

$7 < x < 9$ False

$x > 9$ True

$x < - 4$ True

$- 4 < x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 7$ True

$7 < x < 9$ False

$x > 9$ True

Step 13

The solution consists of all of the true intervals.

$x \leq - 4$ or $- 2 < x \leq 0$ or $0 \leq x < 7$ or $x \geq 9$

Step 14

Combine the intervals.

$x \leq - 4 or - 2 < x < 7 or x \geq 9$

Step 15

Convert the inequality to interval notation.

$(- \infty, - 4] \cup (- 2, 7) \cup [9, \infty)$

Step 16