Precalculus Examples

$p^{2} + 2000 p - 400000 < 0$

Step 1

Convert the inequality to an equation.

$p^{2} + 2000 p - 400000 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 2000$ , and $c = - 400000$ into the quadratic formula and solve for $p$ .

$\frac{- 2000 \pm \sqrt{2000^{2} - 4 \cdot (1 \cdot - 400000)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $2000$ to the power of $2$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot 1 \cdot - 400000}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 400000$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot - 400000}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 400000$ .

$p = \frac{- 2000 \pm \sqrt{4000000 + 1600000}}{2 \cdot 1}$

Step 4.1.3

Add $4000000$ and $1600000$ .

$p = \frac{- 2000 \pm \sqrt{5600000}}{2 \cdot 1}$

Step 4.1.4

Rewrite $5600000$ as $400^{2} \cdot 35$ .

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Step 4.1.4.1

Factor $160000$ out of $5600000$ .

$p = \frac{- 2000 \pm \sqrt{160000 (35)}}{2 \cdot 1}$

Step 4.1.4.2

Rewrite $160000$ as $400^{2}$ .

$p = \frac{- 2000 \pm \sqrt{400^{2} \cdot 35}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2}$

Step 4.3

Simplify $\frac{- 2000 \pm 400 \sqrt{35}}{2}$ .

$p = - 1000 \pm 200 \sqrt{35}$

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $2000$ to the power of $2$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot 1 \cdot - 400000}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 400000$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot - 400000}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 400000$ .

$p = \frac{- 2000 \pm \sqrt{4000000 + 1600000}}{2 \cdot 1}$

Step 5.1.3

Add $4000000$ and $1600000$ .

$p = \frac{- 2000 \pm \sqrt{5600000}}{2 \cdot 1}$

Step 5.1.4

Rewrite $5600000$ as $400^{2} \cdot 35$ .

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Step 5.1.4.1

Factor $160000$ out of $5600000$ .

$p = \frac{- 2000 \pm \sqrt{160000 (35)}}{2 \cdot 1}$

Step 5.1.4.2

Rewrite $160000$ as $400^{2}$ .

$p = \frac{- 2000 \pm \sqrt{400^{2} \cdot 35}}{2 \cdot 1}$

Step 5.1.5

Pull terms out from under the radical.

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2}$

Step 5.3

Simplify $\frac{- 2000 \pm 400 \sqrt{35}}{2}$ .

$p = - 1000 \pm 200 \sqrt{35}$

Step 5.4

Change the $\pm$ to $+$ .

$p = - 1000 + 200 \sqrt{35}$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $2000$ to the power of $2$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot 1 \cdot - 400000}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 400000$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$p = \frac{- 2000 \pm \sqrt{4000000 - 4 \cdot - 400000}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 400000$ .

$p = \frac{- 2000 \pm \sqrt{4000000 + 1600000}}{2 \cdot 1}$

Step 6.1.3

Add $4000000$ and $1600000$ .

$p = \frac{- 2000 \pm \sqrt{5600000}}{2 \cdot 1}$

Step 6.1.4

Rewrite $5600000$ as $400^{2} \cdot 35$ .

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Step 6.1.4.1

Factor $160000$ out of $5600000$ .

$p = \frac{- 2000 \pm \sqrt{160000 (35)}}{2 \cdot 1}$

Step 6.1.4.2

Rewrite $160000$ as $400^{2}$ .

$p = \frac{- 2000 \pm \sqrt{400^{2} \cdot 35}}{2 \cdot 1}$

Step 6.1.5

Pull terms out from under the radical.

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$p = \frac{- 2000 \pm 400 \sqrt{35}}{2}$

Step 6.3

Simplify $\frac{- 2000 \pm 400 \sqrt{35}}{2}$ .

$p = - 1000 \pm 200 \sqrt{35}$

Step 6.4

Change the $\pm$ to $-$ .

$p = - 1000 - 200 \sqrt{35}$

Step 7

Consolidate the solutions.

$p = - 1000 + 200 \sqrt{35}, - 1000 - 200 \sqrt{35}$

Step 8

Use each root to create test intervals.

$p < - 1000 - 200 \sqrt{35}$

$- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$

$p > - 1000 + 200 \sqrt{35}$

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 $p < - 1000 - 200 \sqrt{35}$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $p < - 1000 - 200 \sqrt{35}$ and see if this value makes the original inequality true.

$p = - 2186$

Step 9.1.2

Replace $p$ with $- 2186$ in the original inequality.

${(- 2186)}^{2} + 2000 (- 2186) - 400000 < 0$

Step 9.1.3

The left side $6596$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$ and see if this value makes the original inequality true.

$p = 0$

Step 9.2.2

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

${(0)}^{2} + 2000 (0) - 400000 < 0$

Step 9.2.3

The left side $- 400000$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $p > - 1000 + 200 \sqrt{35}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $p > - 1000 + 200 \sqrt{35}$ and see if this value makes the original inequality true.

$p = 186$

Step 9.3.2

Replace $p$ with $186$ in the original inequality.

${(186)}^{2} + 2000 (186) - 400000 < 0$

Step 9.3.3

The left side $6596$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

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

$p < - 1000 - 200 \sqrt{35}$ False

$- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$ True

$p > - 1000 + 200 \sqrt{35}$ False

$p < - 1000 - 200 \sqrt{35}$ False

$- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$ True

$p > - 1000 + 200 \sqrt{35}$ False

Step 10

The solution consists of all of the true intervals.

$- 1000 - 200 \sqrt{35} < p < - 1000 + 200 \sqrt{35}$

Step 11

Convert the inequality to interval notation.

$(- 1000 - 200 \sqrt{35}, - 1000 + 200 \sqrt{35})$

Step 12