Solve by Completing the Square x^2-x-12=0

$x^{2} - x - 12 = 0$

Add $12$ to both sides of the equation.

$x^{2} - x = 12$

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

${(\frac{b}{2})}^{2} = {(- \frac{1}{2})}^{2}$

Add the term to each side of the equation.

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

Simplify the equation.

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Simplify each term.

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

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Apply the product rule to $- \frac{1}{2}$ .

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

Apply the product rule to $\frac{1}{2}$ .

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

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

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

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

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

One to any power is one.

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

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

$x^{2} - x + \frac{1}{4} = 12 + {(- \frac{1}{2})}^{2}$

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

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Simplify each term.

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

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Apply the product rule to $- \frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = 12 + {(- 1)}^{2} {(\frac{1}{2})}^{2}$

Apply the product rule to $\frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = 12 + {(- 1)}^{2} \frac{1^{2}}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = 12 + 1 \frac{1^{2}}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = 12 + \frac{1^{2}}{2^{2}}$

One to any power is one.

$x^{2} - x + \frac{1}{4} = 12 + \frac{1}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = 12 + \frac{1}{4}$

To write $12$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} - x + \frac{1}{4} = 12 \cdot \frac{4}{4} + \frac{1}{4}$

Combine $12$ and $\frac{4}{4}$ .

$x^{2} - x + \frac{1}{4} = \frac{12 \cdot 4}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x^{2} - x + \frac{1}{4} = \frac{12 \cdot 4 + 1}{4}$

Simplify the numerator.

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Multiply $12$ by $4$ .

$x^{2} - x + \frac{1}{4} = \frac{48 + 1}{4}$

Add $48$ and $1$ .

$x^{2} - x + \frac{1}{4} = \frac{49}{4}$

Factor the perfect trinomial square into ${(x - \frac{1}{2})}^{2}$ .

${(x - \frac{1}{2})}^{2} = \frac{49}{4}$

Solve the equation for $x$ .

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Take the square root of each side of the equation to set up the solution for $x$

${(x - \frac{1}{2})}^{2 \cdot \frac{1}{2}} = \pm \sqrt{\frac{49}{4}}$

Remove the perfect root factor $x - \frac{1}{2}$ under the radical to solve for $x$ .

$x - \frac{1}{2} = \pm \sqrt{\frac{49}{4}}$

Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{49}{4}}$ as $\frac{\sqrt{49}}{\sqrt{4}}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{49}}{\sqrt{4}}$

Simplify the numerator.

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Rewrite $49$ as $7^{2}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{7^{2}}}{\sqrt{4}}$

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

$x - \frac{1}{2} = \pm \frac{7}{\sqrt{4}}$

Simplify the denominator.

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

$x - \frac{1}{2} = \pm \frac{7}{\sqrt{2^{2}}}$

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

$x - \frac{1}{2} = \pm \frac{7}{2}$

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

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

$x - \frac{1}{2} = \frac{7}{2}$

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

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Add $\frac{1}{2}$ to both sides of the equation.

$x = \frac{7}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$x = \frac{7 + 1}{2}$

Add $7$ and $1$ .

$x = \frac{8}{2}$

Divide $8$ by $2$ .

$x = 4$

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

$x - \frac{1}{2} = - \frac{7}{2}$

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

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Add $\frac{1}{2}$ to both sides of the equation.

$x = - \frac{7}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$x = \frac{- 7 + 1}{2}$

Add $- 7$ and $1$ .

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

Divide $- 6$ by $2$ .

$x = - 3$

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

$x = 4, - 3$

Solve by Completing the Square x^2-x-12=0

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