Solve Using the Square Root Property (x-3)^2=-16

${(x - 3)}^{2} = - 16$

Step 1

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

$x - 3 = \pm \sqrt{- 16}$

Step 2

Simplify $\pm \sqrt{- 16}$ .

Rewrite $- 16$ as $- 1 (16)$ .

$x - 3 = \pm \sqrt{- 1 (16)}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x - 3 = \pm \sqrt{- 1} \cdot \sqrt{16}$

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

$x - 3 = \pm i \cdot \sqrt{16}$

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

$x - 3 = \pm i \cdot \sqrt{4^{2}}$

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

$x - 3 = \pm i \cdot 4$

Move $4$ to the left of $i$ .

$x - 3 = \pm 4 i$

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.

$x - 3 = 4 i$

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

Add $3$ to both sides of the equation.

$x = 4 i + 3$

Reorder $4 i$ and $3$ .

$x = 3 + 4 i$

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

$x - 3 = - 4 i$

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

Add $3$ to both sides of the equation.

$x = - 4 i + 3$

Reorder $- 4 i$ and $3$ .

$x = 3 - 4 i$

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

$x = 3 + 4 i, 3 - 4 i$