Algebra Examples

$x^{2} \leq 2 x + 4$

Step 1

Subtract $2 x$ from both sides of the inequality.

$x^{2} - 2 x \leq 4$

Step 2

Subtract $4$ from both sides of the inequality.

$x^{2} - 2 x - 4 \leq 0$

Step 3

Convert the inequality to an equation.

$x^{2} - 2 x - 4 = 0$

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 $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

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

Step 6.1.2

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

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

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

Step 6.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Step 6.1.3

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Step 6.1.4

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

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Step 6.1.4.1

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Step 6.1.4.2

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Step 6.3

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

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

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

Step 7.1.2

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

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Step 7.1.3

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Step 7.1.4

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

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Step 7.1.4.1

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Step 7.1.4.2

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Step 7.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Step 7.3

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

Step 7.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{5}$

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

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

Step 8.1.2

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

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Step 8.1.2.1

Multiply $- 4$ by $1$ .

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

Step 8.1.2.2

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Step 8.1.3

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Step 8.1.4

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

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Step 8.1.4.1

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Step 8.1.4.2

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Step 8.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Step 8.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Step 8.3

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

Step 8.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{5}$

Step 9

Consolidate the solutions.

$x = 1 + \sqrt{5}, 1 - \sqrt{5}$

Step 10

Use each root to create test intervals.

$x < 1 - \sqrt{5}$

$1 - \sqrt{5} < x < 1 + \sqrt{5}$

$x > 1 + \sqrt{5}$

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 $x < 1 - \sqrt{5}$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < 1 - \sqrt{5}$ and see if this value makes the original inequality true.

$x = - 4$

Step 11.1.2

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

${(- 4)}^{2} \leq 2 (- 4) + 4$

Step 11.1.3

The left side $16$ is greater than the right side $- 4$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $1 - \sqrt{5} < x < 1 + \sqrt{5}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $1 - \sqrt{5} < x < 1 + \sqrt{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.2.2

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

${(0)}^{2} \leq 2 (0) + 4$

Step 11.2.3

The left side $0$ is less than the right side $4$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $x > 1 + \sqrt{5}$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > 1 + \sqrt{5}$ and see if this value makes the original inequality true.

$x = 6$

Step 11.3.2

Replace $x$ with $6$ in the original inequality.

${(6)}^{2} \leq 2 (6) + 4$

Step 11.3.3

The left side $36$ is greater than the right side $16$ , which means that the given statement is false.

False

Step 11.4

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

$x < 1 - \sqrt{5}$ False

$1 - \sqrt{5} < x < 1 + \sqrt{5}$ True

$x > 1 + \sqrt{5}$ False

$x < 1 - \sqrt{5}$ False

$1 - \sqrt{5} < x < 1 + \sqrt{5}$ True

$x > 1 + \sqrt{5}$ False

Step 12

The solution consists of all of the true intervals.

$1 - \sqrt{5} \leq x \leq 1 + \sqrt{5}$

Step 13

Convert the inequality to interval notation.

$[1 - \sqrt{5}, 1 + \sqrt{5}]$

Step 14