Trigonometry Examples

$(3 x - 1) (x - 1) < 7$

Step 1

Simplify $(3 x - 1) (x - 1)$ .

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Step 1.1

Expand $(3 x - 1) (x - 1)$ using the FOIL Method.

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Step 1.1.1

Apply the distributive property.

$3 x (x - 1) - 1 (x - 1) < 7$

Step 1.1.2

Apply the distributive property.

$3 x \cdot x + 3 x \cdot - 1 - 1 (x - 1) < 7$

Step 1.1.3

Apply the distributive property.

$3 x \cdot x + 3 x \cdot - 1 - 1 x - 1 \cdot - 1 < 7$

Step 1.2

Simplify and combine like terms.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.1.1.1

Move $x$ .

$3 (x \cdot x) + 3 x \cdot - 1 - 1 x - 1 \cdot - 1 < 7$

Step 1.2.1.1.2

Multiply $x$ by $x$ .

$3 x^{2} + 3 x \cdot - 1 - 1 x - 1 \cdot - 1 < 7$

Step 1.2.1.2

Multiply $- 1$ by $3$ .

$3 x^{2} - 3 x - 1 x - 1 \cdot - 1 < 7$

Step 1.2.1.3

Rewrite $- 1 x$ as $- x$ .

$3 x^{2} - 3 x - x - 1 \cdot - 1 < 7$

Step 1.2.1.4

Multiply $- 1$ by $- 1$ .

$3 x^{2} - 3 x - x + 1 < 7$

Step 1.2.2

Subtract $x$ from $- 3 x$ .

$3 x^{2} - 4 x + 1 < 7$

Step 2

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

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Step 2.1

Subtract $7$ from both sides of the inequality.

$3 x^{2} - 4 x + 1 - 7 < 0$

Step 2.2

Subtract $7$ from $1$ .

$3 x^{2} - 4 x - 6 < 0$

Step 3

Convert the inequality to an equation.

$3 x^{2} - 4 x - 6 = 0$

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 $a = 3$ , $b = - 4$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (3 \cdot - 6)}}{2 \cdot 3}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 3 \cdot - 6}}{2 \cdot 3}$

Step 6.1.2

Multiply $- 4 \cdot 3 \cdot - 6$ .

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Step 6.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{4 \pm \sqrt{16 - 12 \cdot - 6}}{2 \cdot 3}$

Step 6.1.2.2

Multiply $- 12$ by $- 6$ .

$x = \frac{4 \pm \sqrt{16 + 72}}{2 \cdot 3}$

Step 6.1.3

Add $16$ and $72$ .

$x = \frac{4 \pm \sqrt{88}}{2 \cdot 3}$

Step 6.1.4

Rewrite $88$ as $2^{2} \cdot 22$ .

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Step 6.1.4.1

Factor $4$ out of $88$ .

$x = \frac{4 \pm \sqrt{4 (22)}}{2 \cdot 3}$

Step 6.1.4.2

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 22}}{2 \cdot 3}$

Step 6.1.5

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{22}}{2 \cdot 3}$

Step 6.2

Multiply $2$ by $3$ .

$x = \frac{4 \pm 2 \sqrt{22}}{6}$

Step 6.3

Simplify $\frac{4 \pm 2 \sqrt{22}}{6}$ .

$x = \frac{2 \pm \sqrt{22}}{3}$

Step 7

Consolidate the solutions.

$x = \frac{2 + \sqrt{22}}{3}, \frac{2 - \sqrt{22}}{3}$

Step 8

Use each root to create test intervals.

$x < \frac{2 - \sqrt{22}}{3}$

$\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$

$x > \frac{2 + \sqrt{22}}{3}$

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 $x < \frac{2 - \sqrt{22}}{3}$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < \frac{2 - \sqrt{22}}{3}$ and see if this value makes the original inequality true.

$x = - 3$

Step 9.1.2

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

$(3 (- 3) - 1) ((- 3) - 1) < 7$

Step 9.1.3

The left side $40$ is not less than the right side $7$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

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

$(3 (0) - 1) ((0) - 1) < 7$

Step 9.2.3

The left side $1$ is less than the right side $7$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $x > \frac{2 + \sqrt{22}}{3}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > \frac{2 + \sqrt{22}}{3}$ and see if this value makes the original inequality true.

$x = 5$

Step 9.3.2

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

$(3 (5) - 1) ((5) - 1) < 7$

Step 9.3.3

The left side $56$ is not less than the right side $7$ , which means that the given statement is false.

False

Step 9.4

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

$x < \frac{2 - \sqrt{22}}{3}$ False

$\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$ True

$x > \frac{2 + \sqrt{22}}{3}$ False

$x < \frac{2 - \sqrt{22}}{3}$ False

$\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$ True

$x > \frac{2 + \sqrt{22}}{3}$ False

Step 10

The solution consists of all of the true intervals.

$\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$\frac{2 - \sqrt{22}}{3} < x < \frac{2 + \sqrt{22}}{3}$

Interval Notation:

$(\frac{2 - \sqrt{22}}{3}, \frac{2 + \sqrt{22}}{3})$

Step 12