Algebra Examples

$x^{2} \leq 8 x - 13$

Step 1

Subtract $8 x$ from both sides of the inequality.

$x^{2} - 8 x \leq - 13$

Step 2

Add $13$ to both sides of the inequality.

$x^{2} - 8 x + 13 \leq 0$

Step 3

Convert the inequality to an equation.

$x^{2} - 8 x + 13 = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = 1$ , $b = - 8$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot 13$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 13}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $13$ .

$x = \frac{8 \pm \sqrt{64 - 52}}{2 \cdot 1}$

Step 6.1.3

Subtract $52$ from $64$ .

$x = \frac{8 \pm \sqrt{12}}{2 \cdot 1}$

Step 6.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 6.1.4.1

Factor $4$ out of $12$ .

$x = \frac{8 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 6.1.4.2

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

$x = \frac{8 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{3}}{2}$

Step 6.3

Simplify $\frac{8 \pm 2 \sqrt{3}}{2}$ .

$x = 4 \pm \sqrt{3}$

Step 7

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

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot 13$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 13}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $13$ .

$x = \frac{8 \pm \sqrt{64 - 52}}{2 \cdot 1}$

Step 7.1.3

Subtract $52$ from $64$ .

$x = \frac{8 \pm \sqrt{12}}{2 \cdot 1}$

Step 7.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 7.1.4.1

Factor $4$ out of $12$ .

$x = \frac{8 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.1.4.2

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

$x = \frac{8 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.1.5

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{3}}{2}$

Step 7.3

Simplify $\frac{8 \pm 2 \sqrt{3}}{2}$ .

$x = 4 \pm \sqrt{3}$

Step 7.4

Change the $\pm$ to $+$ .

$x = 4 + \sqrt{3}$

Step 8

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

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Step 8.1.2

Multiply $- 4 \cdot 1 \cdot 13$ .

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Step 8.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 13}}{2 \cdot 1}$

Step 8.1.2.2

Multiply $- 4$ by $13$ .

$x = \frac{8 \pm \sqrt{64 - 52}}{2 \cdot 1}$

Step 8.1.3

Subtract $52$ from $64$ .

$x = \frac{8 \pm \sqrt{12}}{2 \cdot 1}$

Step 8.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 8.1.4.1

Factor $4$ out of $12$ .

$x = \frac{8 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 8.1.4.2

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

$x = \frac{8 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 8.1.5

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 8.2

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{3}}{2}$

Step 8.3

Simplify $\frac{8 \pm 2 \sqrt{3}}{2}$ .

$x = 4 \pm \sqrt{3}$

Step 8.4

Change the $\pm$ to $-$ .

$x = 4 - \sqrt{3}$

Step 9

Consolidate the solutions.

$x = 4 + \sqrt{3}, 4 - \sqrt{3}$

Step 10

Use each root to create test intervals.

$x < 4 - \sqrt{3}$

$4 - \sqrt{3} < x < 4 + \sqrt{3}$

$x > 4 + \sqrt{3}$

Step 11

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 11.1

Test a value on the interval $x < 4 - \sqrt{3}$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < 4 - \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.1.2

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

${(0)}^{2} \leq 8 (0) - 13$

Step 11.1.3

The left side $0$ is greater than the right side $- 13$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $4 - \sqrt{3} < x < 4 + \sqrt{3}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $4 - \sqrt{3} < x < 4 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 4$

Step 11.2.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{2} \leq 8 (4) - 13$

Step 11.2.3

The left side $16$ is less than the right side $19$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $x > 4 + \sqrt{3}$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > 4 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 8$

Step 11.3.2

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} \leq 8 (8) - 13$

Step 11.3.3

The left side $64$ is greater than the right side $51$ , which means that the given statement is false.

False

Step 11.4

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

$x < 4 - \sqrt{3}$ False

$4 - \sqrt{3} < x < 4 + \sqrt{3}$ True

$x > 4 + \sqrt{3}$ False

$x < 4 - \sqrt{3}$ False

$4 - \sqrt{3} < x < 4 + \sqrt{3}$ True

$x > 4 + \sqrt{3}$ False

Step 12

The solution consists of all of the true intervals.

$4 - \sqrt{3} \leq x \leq 4 + \sqrt{3}$

Step 13

Convert the inequality to interval notation.

$[4 - \sqrt{3}, 4 + \sqrt{3}]$

Step 14