Solve Using the Square Root Property 3(x-4)^2=27

$3 {(x - 4)}^{2} = 27$

Step 1

Divide each term in $3 {(x - 4)}^{2} = 27$ by $3$ and simplify.

Divide each term in $3 {(x - 4)}^{2} = 27$ by $3$ .

$\frac{3 {(x - 4)}^{2}}{3} = \frac{27}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 {(x - 4)}^{2}}{3} = \frac{27}{3}$

Divide ${(x - 4)}^{2}$ by $1$ .

${(x - 4)}^{2} = \frac{27}{3}$

Simplify the right side.

Divide $27$ by $3$ .

${(x - 4)}^{2} = 9$

Step 2

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

$x - 4 = \pm \sqrt{9}$

Step 3

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

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

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

$x - 4 = \pm 3$

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.

$x - 4 = 3$

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

Add $4$ to both sides of the equation.

$x = 3 + 4$

Add $3$ and $4$ .

$x = 7$

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

$x - 4 = - 3$

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

Add $4$ to both sides of the equation.

$x = - 3 + 4$

Add $- 3$ and $4$ .

$x = 1$

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

$x = 7, 1$