Precalculus Examples

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

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 $x + \frac{1}{x} = 2$ by $x$ to eliminate the fractions.

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Step 2.1

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

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

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 $x$ .

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

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.

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

Step 2.2.1.2.2

Rewrite the expression.

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

Step 3

Solve the equation.

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Step 3.1

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

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

Step 3.2

Factor using the perfect square rule.

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Step 3.2.1

Rearrange terms.

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

Step 3.2.2

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

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

Step 3.2.3

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 3.2.4

Rewrite the polynomial.

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

Step 3.2.5

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 3.3

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

$x - 1 = 0$

Step 3.4

Add $1$ to both sides of the equation.

$x = 1$