Algebra Examples

$\frac{x}{x^{2} + 2 x - 2} \leq 0$

Step 1

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

$x = 0$

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

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

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 4.1.2.2

Multiply $- 4$ by $- 2$ .

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

Step 4.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 1}$

Step 4.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 4.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 4.1.4.2

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{3}}{2}$

Step 4.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{2}$ .

$x = - 1 \pm \sqrt{3}$

Step 5

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{3}, - 1 - \sqrt{3}$

Step 6

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

$x = 0$

$x = - 1 + \sqrt{3}, - 1 - \sqrt{3}$

Step 7

Consolidate the solutions.

$x = 0, - 1 + \sqrt{3}, - 1 - \sqrt{3}$

Step 8

Find the domain of $\frac{x}{x^{2} + 2 x - 2}$ .

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Step 8.1

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

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

Step 8.2

Solve for $x$ .

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Step 8.2.1

Use the quadratic formula to find the solutions.

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

Step 8.2.2

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

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

Step 8.2.3

Simplify.

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Step 8.2.3.1

Simplify the numerator.

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Step 8.2.3.1.1

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

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

Step 8.2.3.1.2

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

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Step 8.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 8.2.3.1.2.2

Multiply $- 4$ by $- 2$ .

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

Step 8.2.3.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 1}$

Step 8.2.3.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 8.2.3.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 8.2.3.1.4.2

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 8.2.3.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 8.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{3}}{2}$

Step 8.2.3.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{2}$ .

$x = - 1 \pm \sqrt{3}$

Step 8.2.4

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{3}, - 1 - \sqrt{3}$

Step 8.3

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

$(- \infty, - 1 - \sqrt{3}) \cup (- 1 - \sqrt{3}, - 1 + \sqrt{3}) \cup (- 1 + \sqrt{3}, \infty)$

Step 9

Use each root to create test intervals.

$x < - 1 - \sqrt{3}$

$- 1 - \sqrt{3} < x < 0$

$0 < x < - 1 + \sqrt{3}$

$x > - 1 + \sqrt{3}$

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 $x < - 1 - \sqrt{3}$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = - 5$

Step 10.1.2

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

$\frac{- 5}{{(- 5)}^{2} + 2 (- 5) - 2} \leq 0$

Step 10.1.3

The left side $- 0.\overline{384615}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 1 - \sqrt{3} < x < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 1 - \sqrt{3} < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 10.2.2

Replace $x$ with $- 1$ in the original inequality.

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

Step 10.2.3

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

False

Step 10.3

Test a value on the interval $0 < x < - 1 + \sqrt{3}$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $0 < x < - 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0.37$

Step 10.3.2

Replace $x$ with $0.37$ in the original inequality.

$\frac{0.37}{{(0.37)}^{2} + 2 (0.37) - 2} \leq 0$

Step 10.3.3

The left side $- 0.32944528$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $x > - 1 + \sqrt{3}$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > - 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 10.4.2

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

$\frac{3}{{(3)}^{2} + 2 (3) - 2} \leq 0$

Step 10.4.3

The left side $0.\overline{230769}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.5

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

$x < - 1 - \sqrt{3}$ True

$- 1 - \sqrt{3} < x < 0$ False

$0 < x < - 1 + \sqrt{3}$ True

$x > - 1 + \sqrt{3}$ False

$x < - 1 - \sqrt{3}$ True

$- 1 - \sqrt{3} < x < 0$ False

$0 < x < - 1 + \sqrt{3}$ True

$x > - 1 + \sqrt{3}$ False

Step 11

The solution consists of all of the true intervals.

$x < - 1 - \sqrt{3}$ or $0 \leq x < - 1 + \sqrt{3}$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 - \sqrt{3} or 0 \leq x < - 1 + \sqrt{3}$

Interval Notation:

$(- \infty, - 1 - \sqrt{3}) \cup [0, - 1 + \sqrt{3})$

Step 13