Precalculus Examples

$x^{2} = \frac{5}{6} x - \frac{1}{6}$

Step 1

Simplify the equation into a proper form to complete the square.

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Step 1.1

Combine $\frac{5}{6}$ and $x$ .

$x^{2} = \frac{5 x}{6} - \frac{1}{6}$

Step 1.2

Subtract $\frac{5 x}{6}$ from both sides of the equation.

$x^{2} - \frac{5 x}{6} = - \frac{1}{6}$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{5}{12})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - \frac{5 x}{6} + {(- \frac{5}{12})}^{2} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.1.1.1.1

Apply the product rule to $- \frac{5}{12}$ .

$x^{2} - \frac{5 x}{6} + {(- 1)}^{2} {(\frac{5}{12})}^{2} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.1.1.1.2

Apply the product rule to $\frac{5}{12}$ .

$x^{2} - \frac{5 x}{6} + {(- 1)}^{2} \frac{5^{2}}{12^{2}} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.1.1.2

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + 1 \frac{5^{2}}{12^{2}} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.1.1.3

Multiply $\frac{5^{2}}{12^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{6} + \frac{5^{2}}{12^{2}} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.1.1.4

Raise $5$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{12^{2}} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.1.1.5

Raise $12$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + {(- \frac{5}{12})}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- \frac{1}{6} + {(- \frac{5}{12})}^{2}$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.2.1.1.1.1

Apply the product rule to $- \frac{5}{12}$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + {(- 1)}^{2} {(\frac{5}{12})}^{2}$

Step 4.2.1.1.1.2

Apply the product rule to $\frac{5}{12}$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + {(- 1)}^{2} \frac{5^{2}}{12^{2}}$

Step 4.2.1.1.2

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + 1 \frac{5^{2}}{12^{2}}$

Step 4.2.1.1.3

Multiply $\frac{5^{2}}{12^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + \frac{5^{2}}{12^{2}}$

Step 4.2.1.1.4

Raise $5$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + \frac{25}{12^{2}}$

Step 4.2.1.1.5

Raise $12$ to the power of $2$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} + \frac{25}{144}$

Step 4.2.1.2

To write $- \frac{1}{6}$ as a fraction with a common denominator, multiply by $\frac{24}{24}$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{1}{6} \cdot \frac{24}{24} + \frac{25}{144}$

Step 4.2.1.3

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.1.3.1

Multiply $\frac{1}{6}$ by $\frac{24}{24}$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{24}{6 \cdot 24} + \frac{25}{144}$

Step 4.2.1.3.2

Multiply $6$ by $24$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = - \frac{24}{144} + \frac{25}{144}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = \frac{- 24 + 25}{144}$

Step 4.2.1.5

Add $- 24$ and $25$ .

$x^{2} - \frac{5 x}{6} + \frac{25}{144} = \frac{1}{144}$

Step 5

Factor the perfect trinomial square into ${(x - \frac{5}{12})}^{2}$ .

${(x - \frac{5}{12})}^{2} = \frac{1}{144}$

Step 6

Solve the equation for $x$ .

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Step 6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{5}{12} = \pm \sqrt{\frac{1}{144}}$

Step 6.2

Simplify $\pm \sqrt{\frac{1}{144}}$ .

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Step 6.2.1

Rewrite $\sqrt{\frac{1}{144}}$ as $\frac{\sqrt{1}}{\sqrt{144}}$ .

$x - \frac{5}{12} = \pm \frac{\sqrt{1}}{\sqrt{144}}$

Step 6.2.2

Any root of $1$ is $1$ .

$x - \frac{5}{12} = \pm \frac{1}{\sqrt{144}}$

Step 6.2.3

Simplify the denominator.

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Step 6.2.3.1

Rewrite $144$ as $12^{2}$ .

$x - \frac{5}{12} = \pm \frac{1}{\sqrt{12^{2}}}$

Step 6.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x - \frac{5}{12} = \pm \frac{1}{12}$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x - \frac{5}{12} = \frac{1}{12}$

Step 6.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.2.1

Add $\frac{5}{12}$ to both sides of the equation.

$x = \frac{1}{12} + \frac{5}{12}$

Step 6.3.2.2

Combine the numerators over the common denominator.

$x = \frac{1 + 5}{12}$

Step 6.3.2.3

Add $1$ and $5$ .

$x = \frac{6}{12}$

Step 6.3.2.4

Cancel the common factor of $6$ and $12$ .

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Step 6.3.2.4.1

Factor $6$ out of $6$ .

$x = \frac{6 (1)}{12}$

Step 6.3.2.4.2

Cancel the common factors.

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Step 6.3.2.4.2.1

Factor $6$ out of $12$ .

$x = \frac{6 \cdot 1}{6 \cdot 2}$

Step 6.3.2.4.2.2

Cancel the common factor.

$x = \frac{6 \cdot 1}{6 \cdot 2}$

Step 6.3.2.4.2.3

Rewrite the expression.

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

Step 6.3.3

Next, use the negative value of the $\pm$ to find the second solution.

$x - \frac{5}{12} = - \frac{1}{12}$

Step 6.3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.4.1

Add $\frac{5}{12}$ to both sides of the equation.

$x = - \frac{1}{12} + \frac{5}{12}$

Step 6.3.4.2

Combine the numerators over the common denominator.

$x = \frac{- 1 + 5}{12}$

Step 6.3.4.3

Add $- 1$ and $5$ .

$x = \frac{4}{12}$

Step 6.3.4.4

Cancel the common factor of $4$ and $12$ .

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Step 6.3.4.4.1

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{12}$

Step 6.3.4.4.2

Cancel the common factors.

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Step 6.3.4.4.2.1

Factor $4$ out of $12$ .

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Step 6.3.4.4.2.2

Cancel the common factor.

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Step 6.3.4.4.2.3

Rewrite the expression.

$x = \frac{1}{3}$

Step 6.3.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{1}{2}, \frac{1}{3}$