Find the Roots (Zeros) f(x)=(x-1)^2-36

$f (x) = {(x - 1)}^{2} - 36$

Step 1

Set ${(x - 1)}^{2} - 36$ equal to $0$ .

${(x - 1)}^{2} - 36 = 0$

Step 2

Solve for $x$ .

Add $36$ to both sides of the equation.

${(x - 1)}^{2} = 36$

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

$x - 1 = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$x - 1 = \pm \sqrt{6^{2}}$

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

$x - 1 = \pm 6$

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.

$x - 1 = 6$

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

Add $1$ to both sides of the equation.

$x = 6 + 1$

Add $6$ and $1$ .

$x = 7$

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

$x - 1 = - 6$

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

Add $1$ to both sides of the equation.

$x = - 6 + 1$

Add $- 6$ and $1$ .

$x = - 5$

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

$x = 7, - 5$

Step 3