Solve by Completing the Square r^2-4r=12

$r^{2} - 4 r = 12$

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 $b$ .

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

Step 2

Add the term to each side of the equation.

$r^{2} - 4 r + {(- 2)}^{2} = 12 + {(- 2)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$r^{2} - 4 r + 4 = 12 + {(- 2)}^{2}$

Simplify the right side.

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

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

$r^{2} - 4 r + 4 = 12 + 4$

Add $12$ and $4$ .

$r^{2} - 4 r + 4 = 16$

Step 4

Factor the perfect trinomial square into ${(r - 2)}^{2}$ .

${(r - 2)}^{2} = 16$

Step 5

Solve the equation for $r$ .

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

$r - 2 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$r - 2 = \pm \sqrt{4^{2}}$

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

$r - 2 = \pm 4$

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.

$r - 2 = 4$

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

Add $2$ to both sides of the equation.

$r = 4 + 2$

Add $4$ and $2$ .

$r = 6$

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

$r - 2 = - 4$

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

Add $2$ to both sides of the equation.

$r = - 4 + 2$

Add $- 4$ and $2$ .

$r = - 2$

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

$r = 6, - 2$

Solve by Completing the Square r^2-4r=12

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