Solve Using the Square Root Property (2x+1)^2+16=0

${(2 x + 1)}^{2} + 16 = 0$

Step 1

Subtract $16$ from both sides of the equation.

${(2 x + 1)}^{2} = - 16$

Step 2

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

$2 x + 1 = \pm \sqrt{- 16}$

Step 3

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

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

$2 x + 1 = \pm \sqrt{- 1 (16)}$

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

$2 x + 1 = \pm \sqrt{- 1} \cdot \sqrt{16}$

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

$2 x + 1 = \pm i \cdot \sqrt{16}$

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

$2 x + 1 = \pm i \cdot \sqrt{4^{2}}$

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

$2 x + 1 = \pm i \cdot 4$

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

$2 x + 1 = \pm 4 i$

Step 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.

$2 x + 1 = 4 i$

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

Subtract $1$ from both sides of the equation.

$2 x = 4 i - 1$

Reorder $4 i$ and $- 1$ .

$2 x = - 1 + 4 i$

Divide each term in $2 x = - 1 + 4 i$ by $2$ and simplify.

Divide each term in $2 x = - 1 + 4 i$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2} + \frac{4 i}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2} + \frac{4 i}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2} + \frac{4 i}{2}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x = - \frac{1}{2} + \frac{4 i}{2}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 i$ .

$x = - \frac{1}{2} + \frac{2 (2 i)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = - \frac{1}{2} + \frac{2 (2 i)}{2 (1)}$

Cancel the common factor.

$x = - \frac{1}{2} + \frac{2 (2 i)}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{1}{2} + \frac{2 i}{1}$

Divide $2 i$ by $1$ .

$x = - \frac{1}{2} + 2 i$

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

$2 x + 1 = - 4 i$

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

Subtract $1$ from both sides of the equation.

$2 x = - 4 i - 1$

Reorder $- 4 i$ and $- 1$ .

$2 x = - 1 - 4 i$

Divide each term in $2 x = - 1 - 4 i$ by $2$ and simplify.

Divide each term in $2 x = - 1 - 4 i$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2} + \frac{- 4 i}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2} + \frac{- 4 i}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2} + \frac{- 4 i}{2}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x = - \frac{1}{2} + \frac{- 4 i}{2}$

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4 i$ .

$x = - \frac{1}{2} + \frac{2 (- 2 i)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = - \frac{1}{2} + \frac{2 (- 2 i)}{2 (1)}$

Cancel the common factor.

$x = - \frac{1}{2} + \frac{2 (- 2 i)}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{1}{2} + \frac{- 2 i}{1}$

Divide $- 2 i$ by $1$ .

$x = - \frac{1}{2} - 2 i$

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

$x = - \frac{1}{2} + 2 i, - \frac{1}{2} - 2 i$

Solve Using the Square Root Property (2x+1)^2+16=0

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