Calculus Examples

$- 8 x^{2} - 8 x + 2 < 9 x - 6$

Step 1

Move all terms containing $x$ to the left side of the inequality.

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Step 1.1

Subtract $9 x$ from both sides of the inequality.

$- 8 x^{2} - 8 x + 2 - 9 x < - 6$

Step 1.2

Subtract $9 x$ from $- 8 x$ .

$- 8 x^{2} - 17 x + 2 < - 6$

Step 2

Move all terms to the left side of the equation and simplify.

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Step 2.1

Add $6$ to both sides of the inequality.

$- 8 x^{2} - 17 x + 2 + 6 < 0$

Step 2.2

Add $2$ and $6$ .

$- 8 x^{2} - 17 x + 8 < 0$

Step 3

Convert the inequality to an equation.

$- 8 x^{2} - 17 x + 8 = 0$

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 $a = - 8$ , $b = - 17$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (- 8 \cdot 8)}}{2 \cdot - 8}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot - 8 \cdot 8}}{2 \cdot - 8}$

Step 6.1.2

Multiply $- 4 \cdot - 8 \cdot 8$ .

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Step 6.1.2.1

Multiply $- 4$ by $- 8$ .

$x = \frac{17 \pm \sqrt{289 + 32 \cdot 8}}{2 \cdot - 8}$

Step 6.1.2.2

Multiply $32$ by $8$ .

$x = \frac{17 \pm \sqrt{289 + 256}}{2 \cdot - 8}$

Step 6.1.3

Add $289$ and $256$ .

$x = \frac{17 \pm \sqrt{545}}{2 \cdot - 8}$

Step 6.2

Multiply $2$ by $- 8$ .

$x = \frac{17 \pm \sqrt{545}}{- 16}$

Step 6.3

Move the negative in front of the fraction.

$x = - \frac{17 \pm \sqrt{545}}{16}$

Step 7

Consolidate the solutions.

$x = - \frac{17 + \sqrt{545}}{16}, - \frac{17 - \sqrt{545}}{16}$

Step 8

Use each root to create test intervals.

$x < - \frac{17 + \sqrt{545}}{16}$

$- \frac{17 + \sqrt{545}}{16} < x < - \frac{17 - \sqrt{545}}{16}$

$x > - \frac{17 - \sqrt{545}}{16}$

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 $x < - \frac{17 + \sqrt{545}}{16}$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - \frac{17 + \sqrt{545}}{16}$ and see if this value makes the original inequality true.

$x = - 5$

Step 9.1.2

Replace $x$ with $- 5$ in the original inequality.

$- 8 {(- 5)}^{2} - 8 \cdot - 5 + 2 < 9 (- 5) - 6$

Step 9.1.3

The left side $- 158$ is less than the right side $- 51$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- \frac{17 + \sqrt{545}}{16} < x < - \frac{17 - \sqrt{545}}{16}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- \frac{17 + \sqrt{545}}{16} < x < - \frac{17 - \sqrt{545}}{16}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

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

$- 8 {(0)}^{2} - 8 \cdot 0 + 2 < 9 (0) - 6$

Step 9.2.3

The left side $2$ is not less than the right side $- 6$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval $x > - \frac{17 - \sqrt{545}}{16}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > - \frac{17 - \sqrt{545}}{16}$ and see if this value makes the original inequality true.

$x = 3$

Step 9.3.2

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

$- 8 {(3)}^{2} - 8 \cdot 3 + 2 < 9 (3) - 6$

Step 9.3.3

The left side $- 94$ is less than the right side $21$ , which means that the given statement is always true.

True

Step 9.4

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

$x < - \frac{17 + \sqrt{545}}{16}$ True

$- \frac{17 + \sqrt{545}}{16} < x < - \frac{17 - \sqrt{545}}{16}$ False

$x > - \frac{17 - \sqrt{545}}{16}$ True

$x < - \frac{17 + \sqrt{545}}{16}$ True

$- \frac{17 + \sqrt{545}}{16} < x < - \frac{17 - \sqrt{545}}{16}$ False

$x > - \frac{17 - \sqrt{545}}{16}$ True

Step 10

The solution consists of all of the true intervals.

$x < - \frac{17 + \sqrt{545}}{16}$ or $x > - \frac{17 - \sqrt{545}}{16}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{17 + \sqrt{545}}{16} or x > - \frac{17 - \sqrt{545}}{16}$

Interval Notation:

$(- \infty, - \frac{17 + \sqrt{545}}{16}) \cup (- \frac{17 - \sqrt{545}}{16}, \infty)$

Step 12