Solve by Completing the Square k^2-8k-48=0

$k^{2} - 8 k - 48 = 0$

Step 1

Add $48$ to both sides of the equation.

$k^{2} - 8 k = 48$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 4)}^{2}$

Step 3

Add the term to each side of the equation.

$k^{2} - 8 k + {(- 4)}^{2} = 48 + {(- 4)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 4$ to the power of $2$ .

$k^{2} - 8 k + 16 = 48 + {(- 4)}^{2}$

Simplify the right side.

Simplify $48 + {(- 4)}^{2}$ .

Raise $- 4$ to the power of $2$ .

$k^{2} - 8 k + 16 = 48 + 16$

Add $48$ and $16$ .

$k^{2} - 8 k + 16 = 64$

Step 5

Factor the perfect trinomial square into ${(k - 4)}^{2}$ .

${(k - 4)}^{2} = 64$

Step 6

Solve the equation for $k$ .

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

$k - 4 = \pm \sqrt{64}$

Simplify $\pm \sqrt{64}$ .

Rewrite $64$ as $8^{2}$ .

$k - 4 = \pm \sqrt{8^{2}}$

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

$k - 4 = \pm 8$

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.

$k - 4 = 8$

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

Add $4$ to both sides of the equation.

$k = 8 + 4$

Add $8$ and $4$ .

$k = 12$

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

$k - 4 = - 8$

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

Add $4$ to both sides of the equation.

$k = - 8 + 4$

Add $- 8$ and $4$ .

$k = - 4$

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

$k = 12, - 4$

Solve by Completing the Square k^2-8k-48=0

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