Solve by Completing the Square x^2-8x+12=0

$x^{2} - 8 x + 12 = 0$

Step 1

Subtract $12$ from both sides of the equation.

$x^{2} - 8 x = - 12$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 4)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 8 x + {(- 4)}^{2} = - 12 + {(- 4)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 4$ to the power of $2$ .

$x^{2} - 8 x + 16 = - 12 + {(- 4)}^{2}$

Simplify the right side.

Simplify $- 12 + {(- 4)}^{2}$ .

Raise $- 4$ to the power of $2$ .

$x^{2} - 8 x + 16 = - 12 + 16$

Add $- 12$ and $16$ .

$x^{2} - 8 x + 16 = 4$

Step 5

Factor the perfect trinomial square into ${(x - 4)}^{2}$ .

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

Step 6

Solve the equation for $x$ .

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

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

Simplify $\pm \sqrt{4}$ .

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

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

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

$x - 4 = \pm 2$

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.

$x - 4 = 2$

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

Add $4$ to both sides of the equation.

$x = 2 + 4$

Add $2$ and $4$ .

$x = 6$

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

$x - 4 = - 2$

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

Add $4$ to both sides of the equation.

$x = - 2 + 4$

Add $- 2$ and $4$ .

$x = 2$

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

$x = 6, 2$

Solve by Completing the Square x^2-8x+12=0

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