Trigonometry Examples

$9 x^{2} - 42 x > - 49$

Step 1

Convert the inequality to an equation.

$9 x^{2} - 42 x = - 49$

Step 2

Add $49$ to both sides of the equation.

$9 x^{2} - 42 x + 49 = 0$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} - 42 x + 49 = 0$

Step 3.2

Rewrite $49$ as $7^{2}$ .

${(3 x)}^{2} - 42 x + 7^{2} = 0$

Step 3.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$42 x = 2 \cdot (3 x) \cdot 7$

Step 3.4

Rewrite the polynomial.

${(3 x)}^{2} - 2 \cdot (3 x) \cdot 7 + 7^{2} = 0$

Step 3.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 x$ and $b = 7$ .

${(3 x - 7)}^{2} = 0$

Step 4

Set the $3 x - 7$ equal to $0$ .

$3 x - 7 = 0$

Step 5

Solve for $x$ .

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Step 5.1

Add $7$ to both sides of the equation.

$3 x = 7$

Step 5.2

Divide each term in $3 x = 7$ by $3$ and simplify.

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Step 5.2.1

Divide each term in $3 x = 7$ by $3$ .

$\frac{3 x}{3} = \frac{7}{3}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{7}{3}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{7}{3}$

Step 6

Use each root to create test intervals.

$x < \frac{7}{3}$

$x > \frac{7}{3}$

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 $x < \frac{7}{3}$ to see if it makes the inequality true.

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Step 7.1.1

Choose a value on the interval $x < \frac{7}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 7.1.2

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

$9 {(0)}^{2} - 42 \cdot 0 > - 49$

Step 7.1.3

The left side $0$ is greater than the right side $- 49$ , which means that the given statement is always true.

True

Step 7.2

Test a value on the interval $x > \frac{7}{3}$ to see if it makes the inequality true.

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Step 7.2.1

Choose a value on the interval $x > \frac{7}{3}$ and see if this value makes the original inequality true.

$x = 5$

Step 7.2.2

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

$9 {(5)}^{2} - 42 \cdot 5 > - 49$

Step 7.2.3

The left side $15$ is greater than the right side $- 49$ , which means that the given statement is always true.

True

Step 7.3

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

$x < \frac{7}{3}$ True

$x > \frac{7}{3}$ True

$x < \frac{7}{3}$ True

$x > \frac{7}{3}$ True

Step 8

The solution consists of all of the true intervals.

$x < \frac{7}{3}$ or $x > \frac{7}{3}$

Step 9

Convert the inequality to interval notation.

$(- \infty, \frac{7}{3}) \cup (\frac{7}{3}, \infty)$

Step 10