Precalculus Examples

$\frac{x^{2} - 4}{x^{2} - 5 x + 6} < 0$

Step 1

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

$x^{2} - 4 = 0$

$x^{2} - 5 x + 6 = 0$

Step 2

Add $4$ to both sides of the equation.

$x^{2} = 4$

Step 3

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

$x = \pm \sqrt{4}$

Step 4

Simplify $\pm \sqrt{4}$ .

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Step 4.1

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

$x = \pm \sqrt{2^{2}}$

Step 4.2

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

$x = \pm 2$

Step 5

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

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Step 5.1

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

$x = 2$

Step 5.2

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

$x = - 2$

Step 5.3

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

$x = 2, - 2$

Step 6

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 6.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 6.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 7

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

$x - 3 = 0$

$x - 2 = 0$

Step 8

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 8.2

Add $3$ to both sides of the equation.

$x = 3$

Step 9

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 9.1

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

$x - 2 = 0$

Step 9.2

Add $2$ to both sides of the equation.

$x = 2$

Step 10

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 11

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

$x = 2, - 2$

$x = 3, 2$

Step 12

Consolidate the solutions.

$x = 2, - 2, 3, 2$

Step 13

Find the domain of $\frac{x^{2} - 4}{x^{2} - 5 x + 6}$ .

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Step 13.1

Set the denominator in $\frac{x^{2} - 4}{x^{2} - 5 x + 6}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x + 6 = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 13.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 13.2.1.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 13.2.2

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

$x - 3 = 0$

$x - 2 = 0$

Step 13.2.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 13.2.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 13.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 13.2.4

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 13.2.4.1

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

$x - 2 = 0$

Step 13.2.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 13.2.5

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 13.3

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

$(- \infty, 2) \cup (2, 3) \cup (3, \infty)$

Step 14

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$2 < x < 3$

$x > 3$

Step 15

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 15.1

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

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Step 15.1.1

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

$x = - 4$

Step 15.1.2

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

$\frac{{(- 4)}^{2} - 4}{{(- 4)}^{2} - 5 \cdot - 4 + 6} < 0$

Step 15.1.3

The left side $0.\overline{285714}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 15.2

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

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Step 15.2.1

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

$x = 0$

Step 15.2.2

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

$\frac{{(0)}^{2} - 4}{{(0)}^{2} - 5 \cdot 0 + 6} < 0$

Step 15.2.3

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

True

Step 15.3

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

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Step 15.3.1

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Step 15.3.2

Replace $x$ with $2.5$ in the original inequality.

$\frac{{(2.5)}^{2} - 4}{{(2.5)}^{2} - 5 \cdot 2.5 + 6} < 0$

Step 15.3.3

The left side $- 9$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 15.4

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 15.4.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 15.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{{(6)}^{2} - 4}{{(6)}^{2} - 5 \cdot 6 + 6} < 0$

Step 15.4.3

The left side $2.\overline{6}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 15.5

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

$x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 3$ True

$x > 3$ False

$x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 3$ True

$x > 3$ False

Step 16

The solution consists of all of the true intervals.

$- 2 < x < 2$ or $2 < x < 3$

Step 17

Convert the inequality to interval notation.

$(- 2, 2) \cup (2, 3)$

Step 18