Solve for k (k-4)^2=-3

${(k - 4)}^{2} = - 3$

Step 1

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

$k - 4 = \pm \sqrt{- 3}$

Step 2

Simplify $\pm \sqrt{- 3}$ .

Rewrite $- 3$ as $- 1 (3)$ .

$k - 4 = \pm \sqrt{- 1 (3)}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$k - 4 = \pm \sqrt{- 1} \cdot \sqrt{3}$

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

$k - 4 = \pm i \sqrt{3}$

Step 3

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.

$k - 4 = i \sqrt{3}$

Add $4$ to both sides of the equation.

$k = i \sqrt{3} + 4$

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

$k - 4 = - i \sqrt{3}$

Add $4$ to both sides of the equation.

$k = - i \sqrt{3} + 4$

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

$k = i \sqrt{3} + 4, - i \sqrt{3} + 4$