Finite Math Examples

$\frac{2 x^{2} - 7 + 3}{{(x - 2)}^{2} (x + 1)} < 0$

Step 1

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

$2 x^{2} - 7 + 3 = 0$

${(x - 2)}^{2} = 0$

$x + 1 = 0$

Step 2

Add $- 7$ and $3$ .

$2 x^{2} - 4 = 0$

Step 3

Add $4$ to both sides of the equation.

$2 x^{2} = 4$

Step 4

Divide each term in $2 x^{2} = 4$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 x^{2} = 4$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{4}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{4}{2}$

Step 4.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{4}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $4$ by $2$ .

$x^{2} = 2$

Step 5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{2}$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{2}$

Step 6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{2}$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{2}, - \sqrt{2}$

Step 7

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 8

Add $2$ to both sides of the equation.

$x = 2$

Step 9

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 10

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

$x = \sqrt{2}, - \sqrt{2}$

$x = 2$

$x = - 1$

Step 11

Consolidate the solutions.

$x = \sqrt{2}, - \sqrt{2}, 2, - 1$

Step 12

Find the domain of $\frac{2 x^{2} - 7 + 3}{{(x - 2)}^{2} (x + 1)}$ .

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Step 12.1

Set the denominator in $\frac{2 x^{2} - 7 + 3}{{(x - 2)}^{2} (x + 1)}$ equal to $0$ to find where the expression is undefined.

${(x - 2)}^{2} (x + 1) = 0$

Step 12.2

Solve for $x$ .

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Step 12.2.1

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

${(x - 2)}^{2} = 0$

$x + 1 = 0$

Step 12.2.2

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 12.2.2.1

Set ${(x - 2)}^{2}$ equal to $0$ .

${(x - 2)}^{2} = 0$

Step 12.2.2.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 12.2.2.2.1

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 12.2.2.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 12.2.3

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

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Step 12.2.3.1

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

$x + 1 = 0$

Step 12.2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 12.2.4

The final solution is all the values that make ${(x - 2)}^{2} (x + 1) = 0$ true.

$x = 2, - 1$

Step 12.3

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

$(- \infty, - 1) \cup (- 1, 2) \cup (2, \infty)$

Step 13

Use each root to create test intervals.

$x < - \sqrt{2}$

$- \sqrt{2} < x < - 1$

$- 1 < x < \sqrt{2}$

$\sqrt{2} < x < 2$

$x > 2$

Step 14

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 14.1

Test a value on the interval $x < - \sqrt{2}$ to see if it makes the inequality true.

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Step 14.1.1

Choose a value on the interval $x < - \sqrt{2}$ and see if this value makes the original inequality true.

$x = - 4$

Step 14.1.2

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

$\frac{2 {(- 4)}^{2} - 7 + 3}{{((- 4) - 2)}^{2} ((- 4) + 1)} < 0$

Step 14.1.3

The left side $- 0.\overline{259}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.2

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

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Step 14.2.1

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

$x = - 1.21$

Step 14.2.2

Replace $x$ with $- 1.21$ in the original inequality.

$\frac{2 {(- 1.21)}^{2} - 7 + 3}{{((- 1.21) - 2)}^{2} ((- 1.21) + 1)} < 0$

Step 14.2.3

The left side $0.49531832$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.3

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

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Step 14.3.1

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

$x = 0$

Step 14.3.2

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

$\frac{2 {(0)}^{2} - 7 + 3}{{((0) - 2)}^{2} ((0) + 1)} < 0$

Step 14.3.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 14.4

Test a value on the interval $\sqrt{2} < x < 2$ to see if it makes the inequality true.

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Step 14.4.1

Choose a value on the interval $\sqrt{2} < x < 2$ and see if this value makes the original inequality true.

$x = 1.71$

Step 14.4.2

Replace $x$ with $1.71$ in the original inequality.

$\frac{2 {(1.71)}^{2} - 7 + 3}{{((1.71) - 2)}^{2} ((1.71) + 1)} < 0$

Step 14.4.3

The left side $8.10930582$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.5

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 14.5.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 14.5.2

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

$\frac{2 {(4)}^{2} - 7 + 3}{{((4) - 2)}^{2} ((4) + 1)} < 0$

Step 14.5.3

The left side $1.4$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 14.6

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

$x < - \sqrt{2}$ True

$- \sqrt{2} < x < - 1$ False

$- 1 < x < \sqrt{2}$ True

$\sqrt{2} < x < 2$ False

$x > 2$ False

$x < - \sqrt{2}$ True

$- \sqrt{2} < x < - 1$ False

$- 1 < x < \sqrt{2}$ True

$\sqrt{2} < x < 2$ False

$x > 2$ False

Step 15

The solution consists of all of the true intervals.

$x < - \sqrt{2}$ or $- 1 < x < \sqrt{2}$

Step 16

The result can be shown in multiple forms.

Inequality Form:

$x < - \sqrt{2} or - 1 < x < \sqrt{2}$

Interval Notation:

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

Step 17