Trigonometry Examples

$3 < 2 y^{2} + 5 y$

Step 1

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

$2 y^{2} + 5 y > 3$

Step 2

Convert the inequality to an equation.

$2 y^{2} + 5 y = 3$

Step 3

Subtract $3$ from both sides of the equation.

$2 y^{2} + 5 y - 3 = 0$

Step 4

Factor by grouping.

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Step 4.1

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

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Step 4.1.1

Factor $5$ out of $5 y$ .

$2 y^{2} + 5 (y) - 3 = 0$

Step 4.1.2

Rewrite $5$ as $- 1$ plus $6$

$2 y^{2} + (- 1 + 6) y - 3 = 0$

Step 4.1.3

Apply the distributive property.

$2 y^{2} - 1 y + 6 y - 3 = 0$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$(2 y^{2} - 1 y) + 6 y - 3 = 0$

Step 4.2.2

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

$y (2 y - 1) + 3 (2 y - 1) = 0$

Step 4.3

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

$(2 y - 1) (y + 3) = 0$

Step 5

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

$2 y - 1 = 0$

$y + 3 = 0$

Step 6

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

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Step 6.1

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

$2 y - 1 = 0$

Step 6.2

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

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Step 6.2.1

Add $1$ to both sides of the equation.

$2 y = 1$

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide $y$ by $1$ .

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

Step 7

Set $y + 3$ equal to $0$ and solve for $y$ .

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Step 7.1

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 8

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

$y = \frac{1}{2}, - 3$

Step 9

Use each root to create test intervals.

$y < - 3$

$- 3 < y < \frac{1}{2}$

$y > \frac{1}{2}$

Step 10

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 10.1

Test a value on the interval $y < - 3$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $y < - 3$ and see if this value makes the original inequality true.

$y = - 6$

Step 10.1.2

Replace $y$ with $- 6$ in the original inequality.

$3 < 2 {(- 6)}^{2} + 5 (- 6)$

Step 10.1.3

The left side $3$ is less than the right side $42$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$y = 0$

Step 10.2.2

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

$3 < 2 {(0)}^{2} + 5 (0)$

Step 10.2.3

The left side $3$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$y = 3$

Step 10.3.2

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

$3 < 2 {(3)}^{2} + 5 (3)$

Step 10.3.3

The left side $3$ is less than the right side $33$ , which means that the given statement is always true.

True

Step 10.4

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

$y < - 3$ True

$- 3 < y < \frac{1}{2}$ False

$y > \frac{1}{2}$ True

$y < - 3$ True

$- 3 < y < \frac{1}{2}$ False

$y > \frac{1}{2}$ True

Step 11

The solution consists of all of the true intervals.

$y < - 3$ or $y > \frac{1}{2}$

Step 12

Convert the inequality to interval notation.

$(- \infty, - 3) \cup (\frac{1}{2}, \infty)$

Step 13