Solve by Factoring (x-4)^2-9=0

${(x - 4)}^{2} - 9 = 0$

Step 1

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

${(x - 4)}^{2} - 3^{2} = 0$

Step 2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x - 4$ and $b = 3$ .

$(x - 4 + 3) (x - 4 - 3) = 0$

Step 3

Simplify.

Add $- 4$ and $3$ .

$(x - 1) (x - 4 - 3) = 0$

Subtract $3$ from $- 4$ .

$(x - 1) (x - 7) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x - 7 = 0$

Step 5

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 6

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 7

The final solution is all the values that make $(x - 1) (x - 7) = 0$ true.

$x = 1, 7$

Solve by Factoring (x-4)^2-9=0

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