Solve the Inequality for x x>=8/(2+x)

$x \geq \frac{8}{2 + x}$

Subtract $\frac{8}{2 + x}$ from both sides of the inequality.

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

Simplify $x - \frac{8}{2 + x}$ .

To write $x$ as a fraction with a common denominator, multiply by $\frac{2 + x}{2 + x}$ .

$\frac{x (2 + x)}{2 + x} - \frac{8}{2 + x} \geq 0$

Combine the numerators over the common denominator.

$\frac{x (2 + x) - 8}{2 + x} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot 2 + x \cdot x - 8}{2 + x} \geq 0$

Move $2$ to the left of $x$ .

$\frac{2 \cdot x + x \cdot x - 8}{2 + x} \geq 0$

Multiply $x$ by $x$ .

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

Reorder terms.

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

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

$- 2, 4$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 4 = 0$

$2 + x = 0$

Add $2$ to both sides of the equation.

$x = 2$

Subtract $4$ from both sides of the equation.

$x = - 4$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = 2$

$x = - 4$

$x = - 2$

Consolidate the solutions.

$x = 2, - 4, - 2$

Find the domain of $\frac{(x - 2) (x + 4)}{2 + x}$ .

Set the denominator in $\frac{(x - 2) (x + 4)}{2 + x}$ equal to $0$ to find where the expression is undefined.

$2 + x = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 2$

$- 2 < x < 2$

$x > 2$

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 $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$- 6 \geq \frac{8}{2 - 6}$

The left side $- 6$ is less than the right side $- 2$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < - 2$ and see if this value makes the original inequality true.

$x = - 3$

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

$- 3 \geq \frac{8}{2 - 3}$

The left side $- 3$ is greater than the right side $- 8$ , 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.

$0 \geq \frac{8}{2 + 0}$

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 $x > 2$ 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.

$4 \geq \frac{8}{2 + 4}$

The left side $4$ is greater than the right side $1.\overline{3}$ , which means that the given statement is always true.

True

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

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

$x < - 4$ False

$- 4 < x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$- 4 \leq x < - 2$ or $x \geq 2$

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x < - 2 or x \geq 2$

Interval Notation:

$[- 4, - 2) \cup [2, \infty)$

Solve the Inequality for x x>=8/(2+x)

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