Trigonometry Examples

$- 16 x^{2} + 32 x - 12 > 0$

Step 1

Convert the inequality to an equation.

$- 16 x^{2} + 32 x - 12 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $- 4$ out of $- 16 x^{2} + 32 x - 12$ .

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Step 2.1.1

Factor $- 4$ out of $- 16 x^{2}$ .

$- 4 (4 x^{2}) + 32 x - 12 = 0$

Step 2.1.2

Factor $- 4$ out of $32 x$ .

$- 4 (4 x^{2}) - 4 (- 8 x) - 12 = 0$

Step 2.1.3

Factor $- 4$ out of $- 12$ .

$- 4 (4 x^{2}) - 4 (- 8 x) - 4 \cdot 3 = 0$

Step 2.1.4

Factor $- 4$ out of $- 4 (4 x^{2}) - 4 (- 8 x)$ .

$- 4 (4 x^{2} - 8 x) - 4 \cdot 3 = 0$

Step 2.1.5

Factor $- 4$ out of $- 4 (4 x^{2} - 8 x) - 4 (3)$ .

$- 4 (4 x^{2} - 8 x + 3) = 0$

Step 2.2

Factor.

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Step 2.2.1

Factor by grouping.

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Step 2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 3 = 12$ and whose sum is $b = - 8$ .

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Step 2.2.1.1.1

Factor $- 8$ out of $- 8 x$ .

$- 4 (4 x^{2} - 8 x + 3) = 0$

Step 2.2.1.1.2

Rewrite $- 8$ as $- 2$ plus $- 6$

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

Step 2.2.1.1.3

Apply the distributive property.

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

Step 2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.1.2.1

Group the first two terms and the last two terms.

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

Step 2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$- 4 (2 x (2 x - 1) - 3 (2 x - 1)) = 0$

Step 2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$- 4 ((2 x - 1) (2 x - 3)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$- 4 (2 x - 1) (2 x - 3) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 1 = 0$

$2 x - 3 = 0$

Step 4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 4.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 4.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 5

Set $2 x - 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Step 5.2

Solve $2 x - 3 = 0$ for $x$ .

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Step 5.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 5.2.2

Divide each term in $2 x = 3$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 6

The final solution is all the values that make $- 4 (2 x - 1) (2 x - 3) = 0$ true.

$x = \frac{1}{2}, \frac{3}{2}$

Step 7

Use each root to create test intervals.

$x < \frac{1}{2}$

$\frac{1}{2} < x < \frac{3}{2}$

$x > \frac{3}{2}$

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 $x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 8.1.1

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

$x = 0$

Step 8.1.2

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

$- 16 {(0)}^{2} + 32 (0) - 12 > 0$

Step 8.1.3

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

False

Step 8.2

Test a value on the interval $\frac{1}{2} < x < \frac{3}{2}$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $\frac{1}{2} < x < \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 8.2.2

Replace $x$ with $1$ in the original inequality.

$- 16 {(1)}^{2} + 32 (1) - 12 > 0$

Step 8.2.3

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

True

Step 8.3

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

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Step 8.3.1

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

$x = 4$

Step 8.3.2

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

$- 16 {(4)}^{2} + 32 (4) - 12 > 0$

Step 8.3.3

The left side $- 140$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 8.4

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

$x < \frac{1}{2}$ False

$\frac{1}{2} < x < \frac{3}{2}$ True

$x > \frac{3}{2}$ False

$x < \frac{1}{2}$ False

$\frac{1}{2} < x < \frac{3}{2}$ True

$x > \frac{3}{2}$ False

Step 9

The solution consists of all of the true intervals.

$\frac{1}{2} < x < \frac{3}{2}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$\frac{1}{2} < x < \frac{3}{2}$

Interval Notation:

$(\frac{1}{2}, \frac{3}{2})$

Step 11