Precalculus Examples

$\frac{2 x^{2} + 2 x + 1}{x (x + 1)} > 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

$2 x^{2} + 2 x + 1 = 0$

$x + 1 = 0$

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 $a = 2$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}$

Step 4.1.2

Multiply $- 4 \cdot 2 \cdot 1$ .

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Step 4.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot 1}}{2 \cdot 2}$

Step 4.1.2.2

Multiply $- 8$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 8}}{2 \cdot 2}$

Step 4.1.3

Subtract $8$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 4}}{2 \cdot 2}$

Step 4.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 2}$

Step 4.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 2}$

Step 4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4}}{2 \cdot 2}$

Step 4.1.7

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

Step 4.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm i \cdot 2}{2 \cdot 2}$

Step 4.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i}{2 \cdot 2}$

Step 4.2

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 2 i}{4}$

Step 4.3

Simplify $\frac{- 2 \pm 2 i}{4}$ .

$x = \frac{- 1 \pm i}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{1}{2} + \frac{i}{2}, - \frac{1}{2} - \frac{i}{2}$

Step 6

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = - \frac{1}{2} + \frac{i}{2}, - \frac{1}{2} - \frac{i}{2}$

$x = - 1$

Step 8

Consolidate the solutions.

$x = 0, - 1$

Step 9

Find the domain of $\frac{2 x^{2} + 2 x + 1}{x (x + 1)}$ .

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Step 9.1

Set the denominator in $\frac{2 x^{2} + 2 x + 1}{x (x + 1)}$ equal to $0$ to find where the expression is undefined.

$x (x + 1) = 0$

Step 9.2

Solve for $x$ .

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Step 9.2.1

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

$x = 0$

$x + 1 = 0$

Step 9.2.2

Set $x$ equal to $0$ .

$x = 0$

Step 9.2.3

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 9.2.3.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 9.2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9.2.4

The final solution is all the values that make $x (x + 1) = 0$ true.

$x = 0, - 1$

Step 9.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 10

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

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 $x < - 1$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 11.1.2

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

$\frac{2 {(- 4)}^{2} + 2 (- 4) + 1}{(- 4) ((- 4) + 1)} > 0$

Step 11.1.3

The left side $2.08\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 11.2.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{2 {(- 0.5)}^{2} + 2 (- 0.5) + 1}{(- 0.5) ((- 0.5) + 1)} > 0$

Step 11.2.3

The left side $- 2$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 11.3.2

Replace $x$ with $2$ in the original inequality.

$\frac{2 {(2)}^{2} + 2 (2) + 1}{(2) ((2) + 1)} > 0$

Step 11.3.3

The left side $2.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.4

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

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

Step 12

The solution consists of all of the true intervals.

$x < - 1$ or $x > 0$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x > 0$

Interval Notation:

$(- \infty, - 1) \cup (0, \infty)$

Step 14