Solve the Inequality for x x-6>2/x

$x - 6 > \frac{2}{x}$

Step 1

Multiply both sides by $x$ .

$(x - 6) x = \frac{2}{x} x$

Step 2

Simplify.

Simplify the left side.

Simplify $(x - 6) x$ .

Apply the distributive property.

$x \cdot x - 6 x = \frac{2}{x} x$

Multiply $x$ by $x$ .

$x^{2} - 6 x = \frac{2}{x} x$

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

$x^{2} - 6 x = \frac{2}{x} x$

Rewrite the expression.

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

Step 3

Solve for $x$ .

Subtract $2$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$x = \frac{6 \pm \sqrt{36 + 8}}{2 \cdot 1}$

Add $36$ and $8$ .

$x = \frac{6 \pm \sqrt{44}}{2 \cdot 1}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$x = \frac{6 \pm \sqrt{4 (11)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{6 \pm \sqrt{2^{2} \cdot 11}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{11}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{11}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{11}}{2}$ .

$x = 3 \pm \sqrt{11}$

The final answer is the combination of both solutions.

$x = 3 + \sqrt{11}, 3 - \sqrt{11}$

Step 4

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

Set the denominator in $\frac{2}{x}$ 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 5

Use each root to create test intervals.

$x < 3 - \sqrt{11}$

$3 - \sqrt{11} < x < 0$

$0 < x < 3 + \sqrt{11}$

$x > 3 + \sqrt{11}$

Step 6

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 $x < 3 - \sqrt{11}$ to see if it makes the inequality true.

Choose a value on the interval $x < 3 - \sqrt{11}$ and see if this value makes the original inequality true.

$x = - 3$

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

$(- 3) - 6 > \frac{2}{- 3}$

The left side $- 9$ is not greater than the right side $- 0.\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $3 - \sqrt{11} < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $3 - \sqrt{11} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.16$

Replace $x$ with $- 0.16$ in the original inequality.

$(- 0.16) - 6 > \frac{2}{- 0.16}$

The left side $- 6.16$ is greater than the right side $- 12.5$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 3 + \sqrt{11}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3 + \sqrt{11}$ and see if this value makes the original inequality true.

$x = 3$

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

$(3) - 6 > \frac{2}{3}$

The left side $- 3$ is not greater than the right side $0.\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $x > 3 + \sqrt{11}$ to see if it makes the inequality true.

Choose a value on the interval $x > 3 + \sqrt{11}$ and see if this value makes the original inequality true.

$x = 9$

Replace $x$ with $9$ in the original inequality.

$(9) - 6 > \frac{2}{9}$

The left side $3$ is greater than the right side $0.\overline{2}$ , which means that the given statement is always true.

True

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

$x < 3 - \sqrt{11}$ False

$3 - \sqrt{11} < x < 0$ True

$0 < x < 3 + \sqrt{11}$ False

$x > 3 + \sqrt{11}$ True

$x < 3 - \sqrt{11}$ False

$3 - \sqrt{11} < x < 0$ True

$0 < x < 3 + \sqrt{11}$ False

$x > 3 + \sqrt{11}$ True

Step 7

The solution consists of all of the true intervals.

$3 - \sqrt{11} < x < 0$ or $x > 3 + \sqrt{11}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$3 - \sqrt{11} < x < 0 or x > 3 + \sqrt{11}$

Interval Notation:

$(3 - \sqrt{11}, 0) \cup (3 + \sqrt{11}, \infty)$

Step 9

Solve the Inequality for x x-6>2/x

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