Solve by Completing the Square x^2-6x+8=0

$x^{2} - 6 x + 8 = 0$

Step 1

Subtract $8$ from both sides of the equation.

$x^{2} - 6 x = - 8$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 3)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 6 x + {(- 3)}^{2} = - 8 + {(- 3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 3$ to the power of $2$ .

$x^{2} - 6 x + 9 = - 8 + {(- 3)}^{2}$

Simplify the right side.

Simplify $- 8 + {(- 3)}^{2}$ .

Raise $- 3$ to the power of $2$ .

$x^{2} - 6 x + 9 = - 8 + 9$

Add $- 8$ and $9$ .

$x^{2} - 6 x + 9 = 1$

Step 5

Factor the perfect trinomial square into ${(x - 3)}^{2}$ .

${(x - 3)}^{2} = 1$

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.

$x - 3 = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x - 3 = \pm 1$

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.

$x - 3 = 1$

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

Add $3$ to both sides of the equation.

$x = 1 + 3$

Add $1$ and $3$ .

$x = 4$

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

$x - 3 = - 1$

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

Add $3$ to both sides of the equation.

$x = - 1 + 3$

Add $- 1$ and $3$ .

$x = 2$

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

$x = 4, 2$

Solve by Completing the Square x^2-6x+8=0

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