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

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

Step 1

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 2

Simplify $\pm \sqrt{16}$ .

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

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

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

$2 x + 1 = \pm 4$

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.

$2 x + 1 = 4$

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

Subtract $1$ from both sides of the equation.

$2 x = 4 - 1$

Subtract $1$ from $4$ .

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

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

$2 x + 1 = - 4$

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

Subtract $1$ from both sides of the equation.

$2 x = - 4 - 1$

Subtract $1$ from $- 4$ .

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

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

$x = \frac{3}{2}, - \frac{5}{2}$

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

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