Precalculus Examples

$\frac{x^{2} - x - 42}{x^{2} - 13 x + 42} > 0$

Step 1

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

$x^{2} - x - 42 = 0$

$x^{2} - 13 x + 42 = 0$

Step 2

Factor $x^{2} - x - 42$ using the AC method.

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Step 2.1

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 $- 42$ and whose sum is $- 1$ .

$- 7, 6$

Step 2.2

Write the factored form using these integers.

$(x - 7) (x + 6) = 0$

Step 3

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

$x - 7 = 0$

$x + 6 = 0$

Step 4

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 5

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 5.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 6

The final solution is all the values that make $(x - 7) (x + 6) = 0$ true.

$x = 7, - 6$

Step 7

Factor $x^{2} - 13 x + 42$ using the AC method.

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Step 7.1

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 $42$ and whose sum is $- 13$ .

$- 7, - 6$

Step 7.2

Write the factored form using these integers.

$(x - 7) (x - 6) = 0$

Step 8

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

$x - 7 = 0$

$x - 6 = 0$

Step 9

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 9.2

Add $7$ to both sides of the equation.

$x = 7$

Step 10

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 10.2

Add $6$ to both sides of the equation.

$x = 6$

Step 11

The final solution is all the values that make $(x - 7) (x - 6) = 0$ true.

$x = 7, 6$

Step 12

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

$x = 7, - 6$

$x = 7, 6$

Step 13

Consolidate the solutions.

$x = 7, - 6, 7, 6$

Step 14

Find the domain of $\frac{x^{2} - x - 42}{x^{2} - 13 x + 42}$ .

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Step 14.1

Set the denominator in $\frac{x^{2} - x - 42}{x^{2} - 13 x + 42}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 13 x + 42 = 0$

Step 14.2

Solve for $x$ .

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Step 14.2.1

Factor $x^{2} - 13 x + 42$ using the AC method.

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Step 14.2.1.1

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 $42$ and whose sum is $- 13$ .

$- 7, - 6$

Step 14.2.1.2

Write the factored form using these integers.

$(x - 7) (x - 6) = 0$

Step 14.2.2

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

$x - 7 = 0$

$x - 6 = 0$

Step 14.2.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 14.2.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 14.2.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 14.2.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 14.2.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 14.2.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 14.2.5

The final solution is all the values that make $(x - 7) (x - 6) = 0$ true.

$x = 7, 6$

Step 14.3

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

$(- \infty, 6) \cup (6, 7) \cup (7, \infty)$

Step 15

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 6$

$6 < x < 7$

$x > 7$

Step 16

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 16.1

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

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Step 16.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 16.1.2

Replace $x$ with $- 8$ in the original inequality.

$\frac{{(- 8)}^{2} - (- 8) - 42}{{(- 8)}^{2} - 13 \cdot - 8 + 42} > 0$

Step 16.1.3

The left side $0.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.2

Test a value on the interval $- 6 < x < 6$ to see if it makes the inequality true.

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Step 16.2.1

Choose a value on the interval $- 6 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

Step 16.2.2

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

$\frac{{(0)}^{2} - (0) - 42}{{(0)}^{2} - 13 \cdot 0 + 42} > 0$

Step 16.2.3

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 16.3

Test a value on the interval $6 < x < 7$ to see if it makes the inequality true.

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Step 16.3.1

Choose a value on the interval $6 < x < 7$ and see if this value makes the original inequality true.

$x = 6.5$

Step 16.3.2

Replace $x$ with $6.5$ in the original inequality.

$\frac{{(6.5)}^{2} - (6.5) - 42}{{(6.5)}^{2} - 13 \cdot 6.5 + 42} > 0$

Step 16.3.3

The left side $25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.4

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 16.4.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 16.4.2

Replace $x$ with $10$ in the original inequality.

$\frac{{(10)}^{2} - (10) - 42}{{(10)}^{2} - 13 \cdot 10 + 42} > 0$

Step 16.4.3

The left side $4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.5

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

$x < - 6$ True

$- 6 < x < 6$ False

$6 < x < 7$ True

$x > 7$ True

$x < - 6$ True

$- 6 < x < 6$ False

$6 < x < 7$ True

$x > 7$ True

Step 17

The solution consists of all of the true intervals.

$x < - 6$ or $6 < x < 7$ or $x > 7$

Step 18

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (6, 7) \cup (7, \infty)$

Step 19