Algebra Examples

$x - 2 > \frac{1}{x}$

Step 1

Solve $1 - \sqrt{2} < x < 0 or x > 1 + \sqrt{2}$ .

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Step 1.1

Multiply both sides by $x$ .

$(x - 2) x = \frac{1}{x} x$

Step 1.2

Simplify.

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Step 1.2.1

Simplify the left side.

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Step 1.2.1.1

Simplify $(x - 2) x$ .

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Step 1.2.1.1.1

Apply the distributive property.

$x \cdot x - 2 x = \frac{1}{x} x$

Step 1.2.1.1.2

Multiply $x$ by $x$ .

$x^{2} - 2 x = \frac{1}{x} x$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Cancel the common factor of $x$ .

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Step 1.2.2.1.1

Cancel the common factor.

$x^{2} - 2 x = \frac{1}{x} x$

Step 1.2.2.1.2

Rewrite the expression.

$x^{2} - 2 x = 1$

Step 1.3

Solve for $x$ .

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Step 1.3.1

Subtract $1$ from both sides of the equation.

$x^{2} - 2 x - 1 = 0$

Step 1.3.2

Use the quadratic formula to find the solutions.

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

Step 1.3.3

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

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

Step 1.3.4

Simplify.

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Step 1.3.4.1

Simplify the numerator.

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Step 1.3.4.1.1

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

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

Step 1.3.4.1.2

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

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Step 1.3.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.3.4.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 1.3.4.1.3

Add $4$ and $4$ .

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

Step 1.3.4.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.3.4.1.4.1

Factor $4$ out of $8$ .

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

Step 1.3.4.1.4.2

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

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

Step 1.3.4.1.5

Pull terms out from under the radical.

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

Step 1.3.4.2

Multiply $2$ by $1$ .

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

Step 1.3.4.3

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

$x = 1 \pm \sqrt{2}$

Step 1.3.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.3.5.1

Simplify the numerator.

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Step 1.3.5.1.1

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

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

Step 1.3.5.1.2

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

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Step 1.3.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.3.5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 1.3.5.1.3

Add $4$ and $4$ .

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

Step 1.3.5.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.3.5.1.4.1

Factor $4$ out of $8$ .

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

Step 1.3.5.1.4.2

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

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

Step 1.3.5.1.5

Pull terms out from under the radical.

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

Step 1.3.5.2

Multiply $2$ by $1$ .

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

Step 1.3.5.3

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

$x = 1 \pm \sqrt{2}$

Step 1.3.5.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{2}$

Step 1.3.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.3.6.1

Simplify the numerator.

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Step 1.3.6.1.1

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

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

Step 1.3.6.1.2

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

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Step 1.3.6.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.3.6.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 1.3.6.1.3

Add $4$ and $4$ .

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

Step 1.3.6.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.3.6.1.4.1

Factor $4$ out of $8$ .

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

Step 1.3.6.1.4.2

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

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

Step 1.3.6.1.5

Pull terms out from under the radical.

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

Step 1.3.6.2

Multiply $2$ by $1$ .

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

Step 1.3.6.3

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

$x = 1 \pm \sqrt{2}$

Step 1.3.6.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{2}$

Step 1.3.7

The final answer is the combination of both solutions.

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 1.4

Find the domain of $x - 2 - \frac{1}{x}$ .

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Step 1.4.1

Set the denominator in $\frac{1}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 1.4.2

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

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

Step 1.5

Use each root to create test intervals.

$x < 1 - \sqrt{2}$

$1 - \sqrt{2} < x < 0$

$0 < x < 1 + \sqrt{2}$

$x > 1 + \sqrt{2}$

Step 1.6

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

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Step 1.6.1

Test a value on the interval $x < 1 - \sqrt{2}$ to see if it makes the inequality true.

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Step 1.6.1.1

Choose a value on the interval $x < 1 - \sqrt{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 1.6.1.2

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

$(- 3) - 2 > \frac{1}{- 3}$

Step 1.6.1.3

The left side $- 5$ is not greater than the right side $- 0.\overline{3}$ , which means that the given statement is false.

False

Step 1.6.2

Test a value on the interval $1 - \sqrt{2} < x < 0$ to see if it makes the inequality true.

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Step 1.6.2.1

Choose a value on the interval $1 - \sqrt{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.21$

Step 1.6.2.2

Replace $x$ with $- 0.21$ in the original inequality.

$(- 0.21) - 2 > \frac{1}{- 0.21}$

Step 1.6.2.3

The left side $- 2.21$ is greater than the right side $- 4.\overline{761904}$ , which means that the given statement is always true.

True

Step 1.6.3

Test a value on the interval $0 < x < 1 + \sqrt{2}$ to see if it makes the inequality true.

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Step 1.6.3.1

Choose a value on the interval $0 < x < 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 1.6.3.2

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

$(1) - 2 > \frac{1}{1}$

Step 1.6.3.3

The left side $- 1$ is not greater than the right side $1$ , which means that the given statement is false.

False

Step 1.6.4

Test a value on the interval $x > 1 + \sqrt{2}$ to see if it makes the inequality true.

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Step 1.6.4.1

Choose a value on the interval $x > 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 5$

Step 1.6.4.2

Replace $x$ with $5$ in the original inequality.

$(5) - 2 > \frac{1}{5}$

Step 1.6.4.3

The left side $3$ is greater than the right side $0.2$ , which means that the given statement is always true.

True

Step 1.6.5

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

$x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 0$ True

$0 < x < 1 + \sqrt{2}$ False

$x > 1 + \sqrt{2}$ True

$x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 0$ True

$0 < x < 1 + \sqrt{2}$ False

$x > 1 + \sqrt{2}$ True

Step 1.7

The solution consists of all of the true intervals.

$1 - \sqrt{2} < x < 0$ or $x > 1 + \sqrt{2}$

Step 2

Use the inequality $1 - \sqrt{2} < x < 0 or x > 1 + \sqrt{2}$ to build the set notation.

${x | 1 - \sqrt{2} < x < 0 or x > 1 + \sqrt{2}}$

Step 3