Precalculus Examples

$\frac{2 x + 3}{x^{2} + 2 x - 15} \geq 0$

Step 1

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

$2 x + 3 = 0$

$x^{2} + 2 x - 15 = 0$

Step 2

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 3

Divide each term in $2 x = - 3$ by $2$ and simplify.

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Step 3.1

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $2$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 4

Factor $x^{2} + 2 x - 15$ using the AC method.

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Step 4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 4.2

Write the factored form using these integers.

$(x - 3) (x + 5) = 0$

Step 5

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

$x - 3 = 0$

$x + 5 = 0$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

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

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Step 7.1

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

$x + 5 = 0$

Step 7.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

The final solution is all the values that make $(x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Step 9

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

$x = - \frac{3}{2}$

$x = 3, - 5$

Step 10

Consolidate the solutions.

$x = - \frac{3}{2}, 3, - 5$

Step 11

Find the domain of $\frac{2 x + 3}{x^{2} + 2 x - 15}$ .

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Step 11.1

Set the denominator in $\frac{2 x + 3}{x^{2} + 2 x - 15}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 15 = 0$

Step 11.2

Solve for $x$ .

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Step 11.2.1

Factor $x^{2} + 2 x - 15$ using the AC method.

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Step 11.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 11.2.1.2

Write the factored form using these integers.

$(x - 3) (x + 5) = 0$

Step 11.2.2

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

$x - 3 = 0$

$x + 5 = 0$

Step 11.2.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 11.2.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 11.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 11.2.4

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

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Step 11.2.4.1

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

$x + 5 = 0$

Step 11.2.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 11.2.5

The final solution is all the values that make $(x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Step 11.3

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

$(- \infty, - 5) \cup (- 5, 3) \cup (3, \infty)$

Step 12

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - \frac{3}{2}$

$- \frac{3}{2} < x < 3$

$x > 3$

Step 13

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 13.1

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

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Step 13.1.1

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

$x = - 8$

Step 13.1.2

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

$\frac{2 (- 8) + 3}{{(- 8)}^{2} + 2 (- 8) - 15} \geq 0$

Step 13.1.3

The left side $- 0.\overline{39}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 13.2

Test a value on the interval $- 5 < x < - \frac{3}{2}$ to see if it makes the inequality true.

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Step 13.2.1

Choose a value on the interval $- 5 < x < - \frac{3}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 13.2.2

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

$\frac{2 (- 3) + 3}{{(- 3)}^{2} + 2 (- 3) - 15} \geq 0$

Step 13.2.3

The left side $0.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.3

Test a value on the interval $- \frac{3}{2} < x < 3$ to see if it makes the inequality true.

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Step 13.3.1

Choose a value on the interval $- \frac{3}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 13.3.2

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

$\frac{2 (0) + 3}{{(0)}^{2} + 2 (0) - 15} \geq 0$

Step 13.3.3

The left side $- 0.2$ is less than the right side $0$ , which means that the given statement is false.

False

Step 13.4

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 13.4.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 13.4.2

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

$\frac{2 (6) + 3}{{(6)}^{2} + 2 (6) - 15} \geq 0$

Step 13.4.3

The left side $0.\overline{45}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 13.5

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

$x < - 5$ False

$- 5 < x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < 3$ False

$x > 3$ True

$x < - 5$ False

$- 5 < x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < 3$ False

$x > 3$ True

Step 14

The solution consists of all of the true intervals.

$- 5 < x \leq - \frac{3}{2}$ or $x > 3$

Step 15

Convert the inequality to interval notation.

$(- 5, - \frac{3}{2}] \cup (3, \infty)$

Step 16