Precalculus Examples

$x^{2} - 6 x - 11 = 0$

Step 1

Add $11$ to both sides of the equation.

$x^{2} - 6 x = 11$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - 6 x + {(- 3)}^{2} = 11 + {(- 3)}^{2}$

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

$x^{2} - 6 x + 9 = 11 + {(- 3)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $11 + {(- 3)}^{2}$ .

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Step 4.2.1.1

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

$x^{2} - 6 x + 9 = 11 + 9$

Step 4.2.1.2

Add $11$ and $9$ .

$x^{2} - 6 x + 9 = 20$

Step 5

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

${(x - 3)}^{2} = 20$

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.

$x - 3 = \pm \sqrt{20}$

Step 6.2

Simplify $\pm \sqrt{20}$ .

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Step 6.2.1

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.2.1.1

Factor $4$ out of $20$ .

$x - 3 = \pm \sqrt{4 (5)}$

Step 6.2.1.2

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

$x - 3 = \pm \sqrt{2^{2} \cdot 5}$

Step 6.2.2

Pull terms out from under the radical.

$x - 3 = \pm 2 \sqrt{5}$

Step 6.3

Add $3$ to both sides of the equation.

$x = \pm 2 \sqrt{5} + 3$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{5} + 3$

Decimal Form:

$x = 7.47213595 \dots, - 1.47213595 \dots$