Algebra Examples

$x + \frac{1}{x} \geq 2$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1$

Step 1.2

The LCM of one and any expression is the expression.

$x$

Step 2

Multiply each term in $x + \frac{1}{x} \geq 2$ by $x$ to eliminate the fractions.

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Step 2.1

Multiply each term in $x + \frac{1}{x} \geq 2$ by $x$ .

$x \cdot x + \frac{1}{x} x \geq 2 x$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $x$ by $x$ .

$x^{2} + \frac{1}{x} x \geq 2 x$

Step 2.2.1.2

Cancel the common factor of $x$ .

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Step 2.2.1.2.1

Cancel the common factor.

$x^{2} + \frac{1}{x} x \geq 2 x$

Step 2.2.1.2.2

Rewrite the expression.

$x^{2} + 1 \geq 2 x$

Step 3

Solve the inequality.

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Step 3.1

Subtract $2 x$ from both sides of the inequality.

$x^{2} + 1 - 2 x \geq 0$

Step 3.2

Convert the inequality to an equation.

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

Step 3.3

Factor using the perfect square rule.

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Step 3.3.1

Rearrange terms.

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

Step 3.3.2

Rewrite $1$ as $1^{2}$ .

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

Step 3.3.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 3.3.4

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 1 + 1^{2} = 0$

Step 3.3.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 3.4

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 3.5

Add $1$ to both sides of the equation.

$x = 1$

Step 4

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

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Step 4.1

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

$x = 0$

Step 4.2

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.

$x < 0$

$0 < x < 1$

$x > 1$

Step 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 6.1

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

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Step 6.1.1

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

$x = - 2$

Step 6.1.2

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

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

Step 6.1.3

The left side $- 2.5$ is less than the right side $2$ , which means that the given statement is false.

False

Step 6.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 6.2.2

Replace $x$ with $0.5$ in the original inequality.

$(0.5) + \frac{1}{0.5} \geq 2$

Step 6.2.3

The left side $2.5$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 6.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 6.3.2

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

$(4) + \frac{1}{4} \geq 2$

Step 6.3.3

The left side $4.25$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 6.4

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

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

Step 7

The solution consists of all of the true intervals.

$0 < x \leq 1$ or $x \geq 1$

Step 8

Combine the intervals.

$x > 0$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x > 0$

Interval Notation:

$(0, \infty)$

Step 10