Algebra Examples

$x^{2} - 12 x + 11 = 0$

Step 1

Subtract $11$ from both sides of the equation.

$x^{2} - 12 x = - 11$

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.

$x^{2} - 12 x + {(- 6)}^{2} = - 11 + {(- 6)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

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

$x^{2} - 12 x + 36 = - 11 + {(- 6)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 11 + {(- 6)}^{2}$ .

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Step 4.2.1.1

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

$x^{2} - 12 x + 36 = - 11 + 36$

Step 4.2.1.2

Add $- 11$ and $36$ .

$x^{2} - 12 x + 36 = 25$

Step 5

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

${(x - 6)}^{2} = 25$

Step 6

Solve the equation for $x$ .

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Step 6.1

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

$x - 6 = \pm \sqrt{25}$

Step 6.2

Simplify $\pm \sqrt{25}$ .

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Step 6.2.1

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

$x - 6 = \pm \sqrt{5^{2}}$

Step 6.2.2

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

$x - 6 = \pm 5$

Step 6.3

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

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Step 6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x - 6 = 5$

Step 6.3.2

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

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Step 6.3.2.1

Add $6$ to both sides of the equation.

$x = 5 + 6$

Step 6.3.2.2

Add $5$ and $6$ .

$x = 11$

Step 6.3.3

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

$x - 6 = - 5$

Step 6.3.4

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

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Step 6.3.4.1

Add $6$ to both sides of the equation.

$x = - 5 + 6$

Step 6.3.4.2

Add $- 5$ and $6$ .

$x = 1$

Step 6.3.5

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

$x = 11, 1$