Solve by Completing the Square c^2+11c=12

$c^{2} + 11 c = 12$

Step 1

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

${(\frac{b}{2})}^{2} = {(\frac{11}{2})}^{2}$

Step 2

Add the term to each side of the equation.

$c^{2} + 11 c + {(\frac{11}{2})}^{2} = 12 + {(\frac{11}{2})}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{11}{2}$ .

$c^{2} + 11 c + \frac{11^{2}}{2^{2}} = 12 + {(\frac{11}{2})}^{2}$

Raise $11$ to the power of $2$ .

$c^{2} + 11 c + \frac{121}{2^{2}} = 12 + {(\frac{11}{2})}^{2}$

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

$c^{2} + 11 c + \frac{121}{4} = 12 + {(\frac{11}{2})}^{2}$

Simplify the right side.

Simplify $12 + {(\frac{11}{2})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{11}{2}$ .

$c^{2} + 11 c + \frac{121}{4} = 12 + \frac{11^{2}}{2^{2}}$

Raise $11$ to the power of $2$ .

$c^{2} + 11 c + \frac{121}{4} = 12 + \frac{121}{2^{2}}$

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

$c^{2} + 11 c + \frac{121}{4} = 12 + \frac{121}{4}$

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

$c^{2} + 11 c + \frac{121}{4} = 12 \cdot \frac{4}{4} + \frac{121}{4}$

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

$c^{2} + 11 c + \frac{121}{4} = \frac{12 \cdot 4}{4} + \frac{121}{4}$

Combine the numerators over the common denominator.

$c^{2} + 11 c + \frac{121}{4} = \frac{12 \cdot 4 + 121}{4}$

Simplify the numerator.

Multiply $12$ by $4$ .

$c^{2} + 11 c + \frac{121}{4} = \frac{48 + 121}{4}$

Add $48$ and $121$ .

$c^{2} + 11 c + \frac{121}{4} = \frac{169}{4}$

Step 4

Factor the perfect trinomial square into ${(c + \frac{11}{2})}^{2}$ .

${(c + \frac{11}{2})}^{2} = \frac{169}{4}$

Step 5

Solve the equation for $c$ .

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

$c + \frac{11}{2} = \pm \sqrt{\frac{169}{4}}$

Simplify $\pm \sqrt{\frac{169}{4}}$ .

Rewrite $\sqrt{\frac{169}{4}}$ as $\frac{\sqrt{169}}{\sqrt{4}}$ .

$c + \frac{11}{2} = \pm \frac{\sqrt{169}}{\sqrt{4}}$

Simplify the numerator.

Rewrite $169$ as $13^{2}$ .

$c + \frac{11}{2} = \pm \frac{\sqrt{13^{2}}}{\sqrt{4}}$

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

$c + \frac{11}{2} = \pm \frac{13}{\sqrt{4}}$

Simplify the denominator.

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

$c + \frac{11}{2} = \pm \frac{13}{\sqrt{2^{2}}}$

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

$c + \frac{11}{2} = \pm \frac{13}{2}$

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

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

$c + \frac{11}{2} = \frac{13}{2}$

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

Subtract $\frac{11}{2}$ from both sides of the equation.

$c = \frac{13}{2} - \frac{11}{2}$

Combine the numerators over the common denominator.

$c = \frac{13 - 11}{2}$

Subtract $11$ from $13$ .

$c = \frac{2}{2}$

Divide $2$ by $2$ .

$c = 1$

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

$c + \frac{11}{2} = - \frac{13}{2}$

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

Subtract $\frac{11}{2}$ from both sides of the equation.

$c = - \frac{13}{2} - \frac{11}{2}$

Combine the numerators over the common denominator.

$c = \frac{- 13 - 11}{2}$

Subtract $11$ from $- 13$ .

$c = \frac{- 24}{2}$

Divide $- 24$ by $2$ .

$c = - 12$

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

$c = 1, - 12$

Solve by Completing the Square c^2+11c=12

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