Find the Roots (Zeros) f(x)=(x-4)^2-25

$f (x) = {(x - 4)}^{2} - 25$

Step 1

Set ${(x - 4)}^{2} - 25$ equal to $0$ .

${(x - 4)}^{2} - 25 = 0$

Step 2

Solve for $x$ .

Add $25$ to both sides of the equation.

${(x - 4)}^{2} = 25$

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

$x - 4 = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

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

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

$x - 4 = \pm 5$

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.

$x - 4 = 5$

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

Add $4$ to both sides of the equation.

$x = 5 + 4$

Add $5$ and $4$ .

$x = 9$

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

$x - 4 = - 5$

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

Add $4$ to both sides of the equation.

$x = - 5 + 4$

Add $- 5$ and $4$ .

$x = - 1$

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

$x = 9, - 1$

Step 3