Solve Using the Square Root Property (2k-5)^2=16

${(2 k - 5)}^{2} = 16$

Step 1

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

$2 k - 5 = \pm \sqrt{16}$

Step 2

Simplify $\pm \sqrt{16}$ .

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

$2 k - 5 = \pm \sqrt{4^{2}}$

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

$2 k - 5 = \pm 4$

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.

$2 k - 5 = 4$

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

Add $5$ to both sides of the equation.

$2 k = 4 + 5$

Add $4$ and $5$ .

$2 k = 9$

Divide each term in $2 k = 9$ by $2$ and simplify.

Divide each term in $2 k = 9$ by $2$ .

$\frac{2 k}{2} = \frac{9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k}{2} = \frac{9}{2}$

Divide $k$ by $1$ .

$k = \frac{9}{2}$

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

$2 k - 5 = - 4$

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

Add $5$ to both sides of the equation.

$2 k = - 4 + 5$

Add $- 4$ and $5$ .

$2 k = 1$

Divide each term in $2 k = 1$ by $2$ and simplify.

Divide each term in $2 k = 1$ by $2$ .

$\frac{2 k}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k}{2} = \frac{1}{2}$

Divide $k$ by $1$ .

$k = \frac{1}{2}$

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

$k = \frac{9}{2}, \frac{1}{2}$

Solve Using the Square Root Property (2k-5)^2=16

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