Convert to Set Notation (x-2)/(x^2-3x-10)<0

$\frac{x - 2}{x^{2} - 3 x - 10} < 0$

Step 1

Solve $x < - 2 or 2 < x < 5$ .

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

$x - 2 = 0$

$x^{2} - 3 x - 10 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Factor $x^{2} - 3 x - 10$ using the AC method.

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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x - 5) (x + 2) = 0$

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

$x - 5 = 0$

$x + 2 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

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

$x = 2$

$x = 5, - 2$

Consolidate the solutions.

$x = 2, 5, - 2$

Find the domain of $\frac{x - 2}{x^{2} - 3 x - 10}$ .

Set the denominator in $\frac{x - 2}{x^{2} - 3 x - 10}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 3 x - 10 = 0$

Solve for $x$ .

Factor $x^{2} - 3 x - 10$ using the AC method.

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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x - 5) (x + 2) = 0$

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

$x - 5 = 0$

$x + 2 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

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

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$2 < x < 5$

$x > 5$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 4$

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

$\frac{(- 4) - 2}{{(- 4)}^{2} - 3 \cdot - 4 - 10} < 0$

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$\frac{(0) - 2}{{(0)}^{2} - 3 \cdot 0 - 10} < 0$

The left side $0.2$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{(4) - 2}{{(4)}^{2} - 3 \cdot 4 - 10} < 0$

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 8$

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

$\frac{(8) - 2}{{(8)}^{2} - 3 \cdot 8 - 10} < 0$

The left side $0.2$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

$x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$x < - 2$ or $2 < x < 5$

Step 2

Use the inequality $x < - 2 or 2 < x < 5$ to build the set notation.

${x | x < - 2 or 2 < x < 5}$

Step 3

Convert to Set Notation (x-2)/(x^2-3x-10)<0

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