Finite Math Examples

$0 > - x^{2} + 7 x + 12$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$- x^{2} + 7 x + 12 < 0$

Step 2

Convert the inequality to an equation.

$- x^{2} + 7 x + 12 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = - 1$ , $b = 7$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 1 \cdot 12)}}{2 \cdot - 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 1 \cdot 12}}{2 \cdot - 1}$

Step 5.1.2

Multiply $- 4 \cdot - 1 \cdot 12$ .

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Step 5.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 7 \pm \sqrt{49 + 4 \cdot 12}}{2 \cdot - 1}$

Step 5.1.2.2

Multiply $4$ by $12$ .

$x = \frac{- 7 \pm \sqrt{49 + 48}}{2 \cdot - 1}$

Step 5.1.3

Add $49$ and $48$ .

$x = \frac{- 7 \pm \sqrt{97}}{2 \cdot - 1}$

Step 5.2

Multiply $2$ by $- 1$ .

$x = \frac{- 7 \pm \sqrt{97}}{- 2}$

Step 5.3

Simplify $\frac{- 7 \pm \sqrt{97}}{- 2}$ .

$x = \frac{7 \pm \sqrt{97}}{2}$

Step 6

Consolidate the solutions.

$x = \frac{7 + \sqrt{97}}{2}, \frac{7 - \sqrt{97}}{2}$

Step 7

Use each root to create test intervals.

$x < \frac{7 - \sqrt{97}}{2}$

$\frac{7 - \sqrt{97}}{2} < x < \frac{7 + \sqrt{97}}{2}$

$x > \frac{7 + \sqrt{97}}{2}$

Step 8

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 8.1

Test a value on the interval $x < \frac{7 - \sqrt{97}}{2}$ to see if it makes the inequality true.

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Step 8.1.1

Choose a value on the interval $x < \frac{7 - \sqrt{97}}{2}$ and see if this value makes the original inequality true.

$x = - 4$

Step 8.1.2

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

$0 > - {(- 4)}^{2} + 7 (- 4) + 12$

Step 8.1.3

The left side $0$ is greater than the right side $- 32$ , which means that the given statement is always true.

True

Step 8.2

Test a value on the interval $\frac{7 - \sqrt{97}}{2} < x < \frac{7 + \sqrt{97}}{2}$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $\frac{7 - \sqrt{97}}{2} < x < \frac{7 + \sqrt{97}}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 8.2.2

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

$0 > - {(0)}^{2} + 7 (0) + 12$

Step 8.2.3

The left side $0$ is not greater than the right side $12$ , which means that the given statement is false.

False

Step 8.3

Test a value on the interval $x > \frac{7 + \sqrt{97}}{2}$ to see if it makes the inequality true.

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Step 8.3.1

Choose a value on the interval $x > \frac{7 + \sqrt{97}}{2}$ and see if this value makes the original inequality true.

$x = 11$

Step 8.3.2

Replace $x$ with $11$ in the original inequality.

$0 > - {(11)}^{2} + 7 (11) + 12$

Step 8.3.3

The left side $0$ is greater than the right side $- 32$ , which means that the given statement is always true.

True

Step 8.4

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

$x < \frac{7 - \sqrt{97}}{2}$ True

$\frac{7 - \sqrt{97}}{2} < x < \frac{7 + \sqrt{97}}{2}$ False

$x > \frac{7 + \sqrt{97}}{2}$ True

$x < \frac{7 - \sqrt{97}}{2}$ True

$\frac{7 - \sqrt{97}}{2} < x < \frac{7 + \sqrt{97}}{2}$ False

$x > \frac{7 + \sqrt{97}}{2}$ True

Step 9

The solution consists of all of the true intervals.

$x < \frac{7 - \sqrt{97}}{2}$ or $x > \frac{7 + \sqrt{97}}{2}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{7 - \sqrt{97}}{2} or x > \frac{7 + \sqrt{97}}{2}$

Interval Notation:

$(- \infty, \frac{7 - \sqrt{97}}{2}) \cup (\frac{7 + \sqrt{97}}{2}, \infty)$

Step 11