Pre-Algebra Examples

$x^{2} - 6 x - 6 < 0$

Step 1

Convert the inequality to an equation.

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 6}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{6 \pm \sqrt{36 + 24}}{2 \cdot 1}$

Step 4.1.3

Add $36$ and $24$ .

$x = \frac{6 \pm \sqrt{60}}{2 \cdot 1}$

Step 4.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 4.1.4.1

Factor $4$ out of $60$ .

$x = \frac{6 \pm \sqrt{4 (15)}}{2 \cdot 1}$

Step 4.1.4.2

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{15}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{15}}{2}$

Step 4.3

Simplify $\frac{6 \pm 2 \sqrt{15}}{2}$ .

$x = 3 \pm \sqrt{15}$

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 6}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{6 \pm \sqrt{36 + 24}}{2 \cdot 1}$

Step 5.1.3

Add $36$ and $24$ .

$x = \frac{6 \pm \sqrt{60}}{2 \cdot 1}$

Step 5.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 5.1.4.1

Factor $4$ out of $60$ .

$x = \frac{6 \pm \sqrt{4 (15)}}{2 \cdot 1}$

Step 5.1.4.2

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 1}$

Step 5.1.5

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{15}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{15}}{2}$

Step 5.3

Simplify $\frac{6 \pm 2 \sqrt{15}}{2}$ .

$x = 3 \pm \sqrt{15}$

Step 5.4

Change the $\pm$ to $+$ .

$x = 3 + \sqrt{15}$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 6$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 6}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 6$ .

$x = \frac{6 \pm \sqrt{36 + 24}}{2 \cdot 1}$

Step 6.1.3

Add $36$ and $24$ .

$x = \frac{6 \pm \sqrt{60}}{2 \cdot 1}$

Step 6.1.4

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 6.1.4.1

Factor $4$ out of $60$ .

$x = \frac{6 \pm \sqrt{4 (15)}}{2 \cdot 1}$

Step 6.1.4.2

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 1}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{15}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{15}}{2}$

Step 6.3

Simplify $\frac{6 \pm 2 \sqrt{15}}{2}$ .

$x = 3 \pm \sqrt{15}$

Step 6.4

Change the $\pm$ to $-$ .

$x = 3 - \sqrt{15}$

Step 7

Consolidate the solutions.

$x = 3 + \sqrt{15}, 3 - \sqrt{15}$

Step 8

Use each root to create test intervals.

$x < 3 - \sqrt{15}$

$3 - \sqrt{15} < x < 3 + \sqrt{15}$

$x > 3 + \sqrt{15}$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 9.1

Test a value on the interval $x < 3 - \sqrt{15}$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < 3 - \sqrt{15}$ and see if this value makes the original inequality true.

$x = - 3$

Step 9.1.2

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} - 6 \cdot - 3 - 6 < 0$

Step 9.1.3

The left side $21$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $3 - \sqrt{15} < x < 3 + \sqrt{15}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $3 - \sqrt{15} < x < 3 + \sqrt{15}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} - 6 \cdot 0 - 6 < 0$

Step 9.2.3

The left side $- 6$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $x > 3 + \sqrt{15}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > 3 + \sqrt{15}$ and see if this value makes the original inequality true.

$x = 9$

Step 9.3.2

Replace $x$ with $9$ in the original inequality.

${(9)}^{2} - 6 \cdot 9 - 6 < 0$

Step 9.3.3

The left side $21$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 3 - \sqrt{15}$ False

$3 - \sqrt{15} < x < 3 + \sqrt{15}$ True

$x > 3 + \sqrt{15}$ False

$x < 3 - \sqrt{15}$ False

$3 - \sqrt{15} < x < 3 + \sqrt{15}$ True

$x > 3 + \sqrt{15}$ False

Step 10

The solution consists of all of the true intervals.

$3 - \sqrt{15} < x < 3 + \sqrt{15}$

Step 11