Solve Using the Square Root Property (3x-1)^2=-16

${(3 x - 1)}^{2} = - 16$

Step 1

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

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

Step 2

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

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

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

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

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

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

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

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

$3 x - 1 = \pm i \cdot \sqrt{4^{2}}$

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

$3 x - 1 = \pm i \cdot 4$

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

$3 x - 1 = \pm 4 i$

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.

$3 x - 1 = 4 i$

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

Add $1$ to both sides of the equation.

$3 x = 4 i + 1$

Reorder $4 i$ and $1$ .

$3 x = 1 + 4 i$

Divide each term in $3 x = 1 + 4 i$ by $3$ and simplify.

Divide each term in $3 x = 1 + 4 i$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3} + \frac{4 i}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3} + \frac{4 i}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3} + \frac{4 i}{3}$

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

$3 x - 1 = - 4 i$

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

Add $1$ to both sides of the equation.

$3 x = - 4 i + 1$

Reorder $- 4 i$ and $1$ .

$3 x = 1 - 4 i$

Divide each term in $3 x = 1 - 4 i$ by $3$ and simplify.

Divide each term in $3 x = 1 - 4 i$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3} + \frac{- 4 i}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3} + \frac{- 4 i}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3} + \frac{- 4 i}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{1}{3} - \frac{4 i}{3}$

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

$x = \frac{1}{3} + \frac{4 i}{3}, \frac{1}{3} - \frac{4 i}{3}$

Solve Using the Square Root Property (3x-1)^2=-16

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