Solve the Inequality for x (x^2+12x)/(x-3)>=4

$\frac{x^{2} + 12 x}{x - 3} \geq 4$

Subtract $4$ from both sides of the inequality.

$\frac{x^{2} + 12 x}{x - 3} - 4 \geq 0$

Simplify $\frac{x^{2} + 12 x}{x - 3} - 4$ .

Factor $x$ out of $x^{2} + 12 x$ .

Factor $x$ out of $x^{2}$ .

$\frac{x \cdot x + 12 x}{x - 3} - 4 \geq 0$

Factor $x$ out of $12 x$ .

$\frac{x \cdot x + x \cdot 12}{x - 3} - 4 \geq 0$

Factor $x$ out of $x \cdot x + x \cdot 12$ .

$\frac{x (x + 12)}{x - 3} - 4 \geq 0$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{x (x + 12)}{x - 3} - 4 \cdot \frac{x - 3}{x - 3} \geq 0$

Combine $- 4$ and $\frac{x - 3}{x - 3}$ .

$\frac{x (x + 12)}{x - 3} + \frac{- 4 (x - 3)}{x - 3} \geq 0$

Combine the numerators over the common denominator.

$\frac{x (x + 12) - 4 (x - 3)}{x - 3} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot x + x \cdot 12 - 4 (x - 3)}{x - 3} \geq 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 12 - 4 (x - 3)}{x - 3} \geq 0$

Move $12$ to the left of $x$ .

$\frac{x^{2} + 12 \cdot x - 4 (x - 3)}{x - 3} \geq 0$

Apply the distributive property.

$\frac{x^{2} + 12 x - 4 x - 4 \cdot - 3}{x - 3} \geq 0$

Multiply $- 4$ by $- 3$ .

$\frac{x^{2} + 12 x - 4 x + 12}{x - 3} \geq 0$

Subtract $4 x$ from $12 x$ .

$\frac{x^{2} + 8 x + 12}{x - 3} \geq 0$

Factor $x^{2} + 8 x + 12$ 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 $12$ and whose sum is $8$ .

$2, 6$

Write the factored form using these integers.

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

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

$x + 2 = 0$

$x + 6 = 0$

$x - 3 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Subtract $6$ from both sides of the equation.

$x = - 6$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = - 2$

$x = - 6$

$x = 3$

Consolidate the solutions.

$x = - 2, - 6, 3$

Find the domain of $\frac{(x + 2) (x + 6)}{x - 3}$ .

Set the denominator in $\frac{(x + 2) (x + 6)}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 2$

$- 2 < x < 3$

$x > 3$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$\frac{{(- 8)}^{2} + 12 (- 8)}{(- 8) - 3} \geq 4$

The left side $2.\overline{90}$ is less than the right side $4$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{{(- 4)}^{2} + 12 (- 4)}{(- 4) - 3} \geq 4$

The left side $4.\overline{571428}$ is greater than the right side $4$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$\frac{{(0)}^{2} + 12 (0)}{(0) - 3} \geq 4$

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

False

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.

$\frac{{(6)}^{2} + 12 (6)}{(6) - 3} \geq 4$

The left side $36$ is greater than the right side $4$ , which means that the given statement is always true.

True

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

$x < - 6$ False

$- 6 < x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 6$ False

$- 6 < x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$- 6 \leq x \leq - 2$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$- 6 \leq x \leq - 2 or x > 3$

Interval Notation:

$[- 6, - 2] \cup (3, \infty)$

Solve the Inequality for x (x^2+12x)/(x-3)>=4

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