Trigonometry Examples

$400000 < - 2 p^{2} + 6000 p$

Step 1

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

$- 2 p^{2} + 6000 p > 400000$

Step 2

Convert the inequality to an equation.

$- 2 p^{2} + 6000 p = 400000$

Step 3

Subtract $400000$ from both sides of the equation.

$- 2 p^{2} + 6000 p - 400000 = 0$

Step 4

Factor $- 2$ out of $- 2 p^{2} + 6000 p - 400000$ .

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Step 4.1

Factor $- 2$ out of $6000 p$ .

$- 2 p^{2} - 2 (- 3000 p) - 400000 = 0$

Step 4.2

Factor $- 2$ out of $- 400000$ .

$- 2 p^{2} - 2 (- 3000 p) - 2 \cdot 200000 = 0$

Step 4.3

Factor $- 2$ out of $- 2 (p^{2}) - 2 (- 3000 p)$ .

$- 2 (p^{2} - 3000 p) - 2 \cdot 200000 = 0$

Step 4.4

Factor $- 2$ out of $- 2 (p^{2} - 3000 p) - 2 (200000)$ .

$- 2 (p^{2} - 3000 p + 200000) = 0$

Step 5

Divide each term in $- 2 (p^{2} - 3000 p + 200000) = 0$ by $- 2$ and simplify.

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Step 5.1

Divide each term in $- 2 (p^{2} - 3000 p + 200000) = 0$ by $- 2$ .

$\frac{- 2 (p^{2} - 3000 p + 200000)}{- 2} = \frac{0}{- 2}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $- 2$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{- 2 (p^{2} - 3000 p + 200000)}{- 2} = \frac{0}{- 2}$

Step 5.2.1.2

Divide $p^{2} - 3000 p + 200000$ by $1$ .

$p^{2} - 3000 p + 200000 = \frac{0}{- 2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $0$ by $- 2$ .

$p^{2} - 3000 p + 200000 = 0$

Step 6

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7

Substitute the values $a = 1$ , $b = - 3000$ , and $c = 200000$ into the quadratic formula and solve for $p$ .

$\frac{3000 \pm \sqrt{{(- 3000)}^{2} - 4 \cdot (1 \cdot 200000)}}{2 \cdot 1}$

Step 8

Simplify.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Raise $- 3000$ to the power of $2$ .

$p = \frac{3000 \pm \sqrt{9000000 - 4 \cdot 1 \cdot 200000}}{2 \cdot 1}$

Step 8.1.2

Multiply $- 4 \cdot 1 \cdot 200000$ .

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Step 8.1.2.1

Multiply $- 4$ by $1$ .

$p = \frac{3000 \pm \sqrt{9000000 - 4 \cdot 200000}}{2 \cdot 1}$

Step 8.1.2.2

Multiply $- 4$ by $200000$ .

$p = \frac{3000 \pm \sqrt{9000000 - 800000}}{2 \cdot 1}$

Step 8.1.3

Subtract $800000$ from $9000000$ .

$p = \frac{3000 \pm \sqrt{8200000}}{2 \cdot 1}$

Step 8.1.4

Rewrite $8200000$ as $200^{2} \cdot 205$ .

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Step 8.1.4.1

Factor $40000$ out of $8200000$ .

$p = \frac{3000 \pm \sqrt{40000 (205)}}{2 \cdot 1}$

Step 8.1.4.2

Rewrite $40000$ as $200^{2}$ .

$p = \frac{3000 \pm \sqrt{200^{2} \cdot 205}}{2 \cdot 1}$

Step 8.1.5

Pull terms out from under the radical.

$p = \frac{3000 \pm 200 \sqrt{205}}{2 \cdot 1}$

Step 8.2

Multiply $2$ by $1$ .

$p = \frac{3000 \pm 200 \sqrt{205}}{2}$

Step 8.3

Simplify $\frac{3000 \pm 200 \sqrt{205}}{2}$ .

$p = 1500 \pm 100 \sqrt{205}$

Step 9

The final answer is the combination of both solutions.

$p = 1500 + 100 \sqrt{205}, 1500 - 100 \sqrt{205}$

Step 10

Use each root to create test intervals.

$p < 1500 - 100 \sqrt{205}$

$1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$

$p > 1500 + 100 \sqrt{205}$

Step 11

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 11.1

Test a value on the interval $p < 1500 - 100 \sqrt{205}$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $p < 1500 - 100 \sqrt{205}$ and see if this value makes the original inequality true.

$p = 0$

Step 11.1.2

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

$400000 < - 2 {(0)}^{2} + 6000 (0)$

Step 11.1.3

The left side $400000$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$ and see if this value makes the original inequality true.

$p = 71$

Step 11.2.2

Replace $p$ with $71$ in the original inequality.

$400000 < - 2 {(71)}^{2} + 6000 (71)$

Step 11.2.3

The left side $400000$ is less than the right side $415918$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $p > 1500 + 100 \sqrt{205}$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $p > 1500 + 100 \sqrt{205}$ and see if this value makes the original inequality true.

$p = 2934$

Step 11.3.2

Replace $p$ with $2934$ in the original inequality.

$400000 < - 2 {(2934)}^{2} + 6000 (2934)$

Step 11.3.3

The left side $400000$ is not less than the right side $387288$ , which means that the given statement is false.

False

Step 11.4

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

$p < 1500 - 100 \sqrt{205}$ False

$1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$ True

$p > 1500 + 100 \sqrt{205}$ False

$p < 1500 - 100 \sqrt{205}$ False

$1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$ True

$p > 1500 + 100 \sqrt{205}$ False

Step 12

The solution consists of all of the true intervals.

$1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$1500 - 100 \sqrt{205} < p < 1500 + 100 \sqrt{205}$

Interval Notation:

$(1500 - 100 \sqrt{205}, 1500 + 100 \sqrt{205})$

Step 14