Solve the Inequality for x (x^2+x-6)/x>=0

$\frac{x^{2} + x - 6}{x} \geq 0$

Step 1

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

$x = 0$

$x^{2} + x - 6 = 0$

Step 2

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$(x - 2) (x + 3) = 0$

Step 3

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

$x - 2 = 0$

$x + 3 = 0$

Step 4

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 7

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

$x = 0$

$x = 2, - 3$

Step 8

Consolidate the solutions.

$x = 0, 2, - 3$

Step 9

Find the domain of $\frac{x^{2} + x - 6}{x}$ .

Set the denominator in $\frac{x^{2} + x - 6}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

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

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

Step 10

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$0 < x < 2$

$x > 2$

Step 11

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

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

The left side $- 4$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

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

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 1$

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

$\frac{{(1)}^{2} + 1 - 6}{1} \geq 0$

The left side $- 4$ is less than the right side $0$ , which means that the given statement is false.

False

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.

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

The left side $3.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 3$ False

$- 3 < x < 0$ True

$0 < x < 2$ False

$x > 2$ True

$x < - 3$ False

$- 3 < x < 0$ True

$0 < x < 2$ False

$x > 2$ True

Step 12

The solution consists of all of the true intervals.

$- 3 \leq x < 0$ or $x \geq 2$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x < 0 or x \geq 2$

Interval Notation:

$[- 3, 0) \cup [2, \infty)$

Step 14

Solve the Inequality for x (x^2+x-6)/x>=0

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