Solve by Completing the Square x^2-12x+27=0

$x^{2} - 12 x + 27 = 0$

Step 1

Subtract $27$ from both sides of the equation.

$x^{2} - 12 x = - 27$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 6)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 12 x + {(- 6)}^{2} = - 27 + {(- 6)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 6$ to the power of $2$ .

$x^{2} - 12 x + 36 = - 27 + {(- 6)}^{2}$

Simplify the right side.

Simplify $- 27 + {(- 6)}^{2}$ .

Raise $- 6$ to the power of $2$ .

$x^{2} - 12 x + 36 = - 27 + 36$

Add $- 27$ and $36$ .

$x^{2} - 12 x + 36 = 9$

Step 5

Factor the perfect trinomial square into ${(x - 6)}^{2}$ .

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

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.

$x - 6 = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$x - 6 = \pm \sqrt{3^{2}}$

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

$x - 6 = \pm 3$

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.

$x - 6 = 3$

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

Add $6$ to both sides of the equation.

$x = 3 + 6$

Add $3$ and $6$ .

$x = 9$

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

$x - 6 = - 3$

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

Add $6$ to both sides of the equation.

$x = - 3 + 6$

Add $- 3$ and $6$ .

$x = 3$

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

$x = 9, 3$

Solve by Completing the Square x^2-12x+27=0

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