Finite Math Examples

$a^{3} - 25 a < 0$

Step 1

Convert the inequality to an equation.

$a^{3} - 25 a = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $a$ out of $a^{3} - 25 a$ .

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Step 2.1.1

Factor $a$ out of $a^{3}$ .

$a \cdot a^{2} - 25 a = 0$

Step 2.1.2

Factor $a$ out of $- 25 a$ .

$a \cdot a^{2} + a \cdot - 25 = 0$

Step 2.1.3

Factor $a$ out of $a \cdot a^{2} + a \cdot - 25$ .

$a (a^{2} - 25) = 0$

Step 2.2

Rewrite $25$ as $5^{2}$ .

$a (a^{2} - 5^{2}) = 0$

Step 2.3

Factor.

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Step 2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = a$ and $b = 5$ .

$a ((a + 5) (a - 5)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$a (a + 5) (a - 5) = 0$

Step 3

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

$a = 0$

$a + 5 = 0$

$a - 5 = 0$

Step 4

Set $a$ equal to $0$ .

$a = 0$

Step 5

Set $a + 5$ equal to $0$ and solve for $a$ .

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Step 5.1

Set $a + 5$ equal to $0$ .

$a + 5 = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$a = - 5$

Step 6

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

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Step 6.1

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

$a - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$a = 5$

Step 7

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

$a = 0, - 5, 5$

Step 8

Use each root to create test intervals.

$a < - 5$

$- 5 < a < 0$

$0 < a < 5$

$a > 5$

Step 9

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 9.1

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

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Step 9.1.1

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

$a = - 8$

Step 9.1.2

Replace $a$ with $- 8$ in the original inequality.

${(- 8)}^{3} - 25 \cdot - 8 < 0$

Step 9.1.3

The left side $- 312$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

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

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Step 9.2.1

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

$a = - 2$

Step 9.2.2

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

${(- 2)}^{3} - 25 \cdot - 2 < 0$

Step 9.2.3

The left side $42$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.3

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

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Step 9.3.1

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

$a = 2$

Step 9.3.2

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

${(2)}^{3} - 25 \cdot 2 < 0$

Step 9.3.3

The left side $- 42$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.4

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

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Step 9.4.1

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

$a = 8$

Step 9.4.2

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

${(8)}^{3} - 25 \cdot 8 < 0$

Step 9.4.3

The left side $312$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.5

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

$a < - 5$ True

$- 5 < a < 0$ False

$0 < a < 5$ True

$a > 5$ False

$a < - 5$ True

$- 5 < a < 0$ False

$0 < a < 5$ True

$a > 5$ False

Step 10

The solution consists of all of the true intervals.

$a < - 5$ or $0 < a < 5$

Step 11

Convert the inequality to interval notation.

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

Step 12