Solve by Completing the Square k^2-4k+1=-5

$k^{2} - 4 k + 1 = - 5$

Step 1

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

Subtract $1$ from both sides of the equation.

$k^{2} - 4 k = - 5 - 1$

Subtract $1$ from $- 5$ .

$k^{2} - 4 k = - 6$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$k^{2} - 4 k + {(- 2)}^{2} = - 6 + {(- 2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$k^{2} - 4 k + 4 = - 6 + {(- 2)}^{2}$

Simplify the right side.

Simplify $- 6 + {(- 2)}^{2}$ .

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

$k^{2} - 4 k + 4 = - 6 + 4$

Add $- 6$ and $4$ .

$k^{2} - 4 k + 4 = - 2$

Step 5

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

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

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.

$k - 2 = \pm \sqrt{- 2}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$k - 2 = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$k - 2 = \pm \sqrt{- 1} \cdot \sqrt{2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$k - 2 = \pm i \sqrt{2}$

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

Add $2$ to both sides of the equation.

$k = \pm i \sqrt{2} + 2$

Reorder $\pm i \sqrt{2}$ and $2$ .

$k = 2 \pm i \sqrt{2}$