Finite Math Examples

$4 x^{2} + 4 x - 1 > 0$

Step 1

Convert the inequality to an equation.

$4 x^{2} + 4 x - 1 = 0$

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

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $4$ to the power of $2$ .

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

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $4$ .

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

Step 4.1.2.2

Multiply $- 16$ by $- 1$ .

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

Step 4.1.3

Add $16$ and $16$ .

$x = \frac{- 4 \pm \sqrt{32}}{2 \cdot 4}$

Step 4.1.4

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 4.1.4.1

Factor $16$ out of $32$ .

$x = \frac{- 4 \pm \sqrt{16 (2)}}{2 \cdot 4}$

Step 4.1.4.2

Rewrite $16$ as $4^{2}$ .

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

Step 4.1.5

Pull terms out from under the radical.

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

Step 4.2

Multiply $2$ by $4$ .

$x = \frac{- 4 \pm 4 \sqrt{2}}{8}$

Step 4.3

Simplify $\frac{- 4 \pm 4 \sqrt{2}}{8}$ .

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

Step 5

Consolidate the solutions.

$x = - \frac{1 - \sqrt{2}}{2}, - \frac{1 + \sqrt{2}}{2}$

Step 6

Use each root to create test intervals.

$x < - \frac{1 + \sqrt{2}}{2}$

$- \frac{1 + \sqrt{2}}{2} < x < - \frac{1 - \sqrt{2}}{2}$

$x > - \frac{1 - \sqrt{2}}{2}$

Step 7

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 7.1

Test a value on the interval $x < - \frac{1 + \sqrt{2}}{2}$ to see if it makes the inequality true.

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Step 7.1.1

Choose a value on the interval $x < - \frac{1 + \sqrt{2}}{2}$ and see if this value makes the original inequality true.

$x = - 4$

Step 7.1.2

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

$4 {(- 4)}^{2} + 4 (- 4) - 1 > 0$

Step 7.1.3

The left side $47$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 7.2

Test a value on the interval $- \frac{1 + \sqrt{2}}{2} < x < - \frac{1 - \sqrt{2}}{2}$ to see if it makes the inequality true.

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Step 7.2.1

Choose a value on the interval $- \frac{1 + \sqrt{2}}{2} < x < - \frac{1 - \sqrt{2}}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 7.2.2

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

$4 {(0)}^{2} + 4 (0) - 1 > 0$

Step 7.2.3

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 7.3

Test a value on the interval $x > - \frac{1 - \sqrt{2}}{2}$ to see if it makes the inequality true.

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Step 7.3.1

Choose a value on the interval $x > - \frac{1 - \sqrt{2}}{2}$ and see if this value makes the original inequality true.

$x = 3$

Step 7.3.2

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

$4 {(3)}^{2} + 4 (3) - 1 > 0$

Step 7.3.3

The left side $47$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 7.4

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

$x < - \frac{1 + \sqrt{2}}{2}$ True

$- \frac{1 + \sqrt{2}}{2} < x < - \frac{1 - \sqrt{2}}{2}$ False

$x > - \frac{1 - \sqrt{2}}{2}$ True

$x < - \frac{1 + \sqrt{2}}{2}$ True

$- \frac{1 + \sqrt{2}}{2} < x < - \frac{1 - \sqrt{2}}{2}$ False

$x > - \frac{1 - \sqrt{2}}{2}$ True

Step 8

The solution consists of all of the true intervals.

$x < - \frac{1 + \sqrt{2}}{2}$ or $x > - \frac{1 - \sqrt{2}}{2}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{1 + \sqrt{2}}{2} or x > - \frac{1 - \sqrt{2}}{2}$

Interval Notation:

$(- \infty, - \frac{1 + \sqrt{2}}{2}) \cup (- \frac{1 - \sqrt{2}}{2}, \infty)$

Step 10