Precalculus Examples

$\frac{x^{2} - 9 x + 18}{4 x^{2} - 25} \leq 0$

Step 1

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

$x^{2} - 9 x + 18 = 0$

$4 x^{2} - 25 = 0$

Step 2

Factor $x^{2} - 9 x + 18$ 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 $18$ and whose sum is $- 9$ .

$- 6, - 3$

Step 2.2

Write the factored form using these integers.

$(x - 6) (x - 3) = 0$

Step 3

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

$x - 6 = 0$

$x - 3 = 0$

Step 4

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

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Step 4.1

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

$x - 6 = 0$

Step 4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 5

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

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Step 5.1

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

$x - 3 = 0$

Step 5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6

The final solution is all the values that make $(x - 6) (x - 3) = 0$ true.

$x = 6, 3$

Step 7

Add $25$ to both sides of the equation.

$4 x^{2} = 25$

Step 8

Divide each term in $4 x^{2} = 25$ by $4$ and simplify.

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Step 8.1

Divide each term in $4 x^{2} = 25$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{25}{4}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $4$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{25}{4}$

Step 8.2.1.2

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

$x^{2} = \frac{25}{4}$

Step 9

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

$x = \pm \sqrt{\frac{25}{4}}$

Step 10

Simplify $\pm \sqrt{\frac{25}{4}}$ .

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Step 10.1

Rewrite $\sqrt{\frac{25}{4}}$ as $\frac{\sqrt{25}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{25}}{\sqrt{4}}$

Step 10.2

Simplify the numerator.

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Step 10.2.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \frac{\sqrt{5^{2}}}{\sqrt{4}}$

Step 10.2.2

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

$x = \pm \frac{5}{\sqrt{4}}$

Step 10.3

Simplify the denominator.

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Step 10.3.1

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

$x = \pm \frac{5}{\sqrt{2^{2}}}$

Step 10.3.2

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

$x = \pm \frac{5}{2}$

Step 11

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

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Step 11.1

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

$x = \frac{5}{2}$

Step 11.2

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

$x = - \frac{5}{2}$

Step 11.3

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

$x = \frac{5}{2}, - \frac{5}{2}$

Step 12

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

$x = 6, 3$

$x = \frac{5}{2}, - \frac{5}{2}$

Step 13

Consolidate the solutions.

$x = 6, 3, \frac{5}{2}, - \frac{5}{2}$

Step 14

Find the domain of $\frac{x^{2} - 9 x + 18}{4 x^{2} - 25}$ .

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Step 14.1

Set the denominator in $\frac{x^{2} - 9 x + 18}{4 x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$4 x^{2} - 25 = 0$

Step 14.2

Solve for $x$ .

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Step 14.2.1

Add $25$ to both sides of the equation.

$4 x^{2} = 25$

Step 14.2.2

Divide each term in $4 x^{2} = 25$ by $4$ and simplify.

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Step 14.2.2.1

Divide each term in $4 x^{2} = 25$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{25}{4}$

Step 14.2.2.2

Simplify the left side.

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Step 14.2.2.2.1

Cancel the common factor of $4$ .

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Step 14.2.2.2.1.1

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{25}{4}$

Step 14.2.2.2.1.2

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

$x^{2} = \frac{25}{4}$

Step 14.2.3

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

$x = \pm \sqrt{\frac{25}{4}}$

Step 14.2.4

Simplify $\pm \sqrt{\frac{25}{4}}$ .

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Step 14.2.4.1

Rewrite $\sqrt{\frac{25}{4}}$ as $\frac{\sqrt{25}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{25}}{\sqrt{4}}$

Step 14.2.4.2

Simplify the numerator.

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Step 14.2.4.2.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \frac{\sqrt{5^{2}}}{\sqrt{4}}$

Step 14.2.4.2.2

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

$x = \pm \frac{5}{\sqrt{4}}$

Step 14.2.4.3

Simplify the denominator.

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Step 14.2.4.3.1

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

$x = \pm \frac{5}{\sqrt{2^{2}}}$

Step 14.2.4.3.2

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

$x = \pm \frac{5}{2}$

Step 14.2.5

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

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Step 14.2.5.1

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

$x = \frac{5}{2}$

Step 14.2.5.2

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

$x = - \frac{5}{2}$

Step 14.2.5.3

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

$x = \frac{5}{2}, - \frac{5}{2}$

Step 14.3

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

$(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$

Step 15

Use each root to create test intervals.

$x < - \frac{5}{2}$

$- \frac{5}{2} < x < \frac{5}{2}$

$\frac{5}{2} < x < 3$

$3 < x < 6$

$x > 6$

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 $x < - \frac{5}{2}$ to see if it makes the inequality true.

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Step 16.1.1

Choose a value on the interval $x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Step 16.1.2

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

$\frac{{(- 5)}^{2} - 9 \cdot - 5 + 18}{4 {(- 5)}^{2} - 25} \leq 0$

Step 16.1.3

The left side $1.17\overline{3}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 16.2

Test a value on the interval $- \frac{5}{2} < x < \frac{5}{2}$ to see if it makes the inequality true.

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Step 16.2.1

Choose a value on the interval $- \frac{5}{2} < x < \frac{5}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 16.2.2

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

$\frac{{(0)}^{2} - 9 \cdot 0 + 18}{4 {(0)}^{2} - 25} \leq 0$

Step 16.2.3

The left side $- 0.72$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 16.3

Test a value on the interval $\frac{5}{2} < x < 3$ to see if it makes the inequality true.

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Step 16.3.1

Choose a value on the interval $\frac{5}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 2.75$

Step 16.3.2

Replace $x$ with $2.75$ in the original inequality.

$\frac{{(2.75)}^{2} - 9 \cdot 2.75 + 18}{4 {(2.75)}^{2} - 25} \leq 0$

Step 16.3.3

The left side $0.1547619$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 16.4

Test a value on the interval $3 < x < 6$ to see if it makes the inequality true.

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Step 16.4.1

Choose a value on the interval $3 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

Step 16.4.2

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

$\frac{{(4)}^{2} - 9 \cdot 4 + 18}{4 {(4)}^{2} - 25} \leq 0$

Step 16.4.3

The left side $- 0.\overline{051282}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 16.5

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

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Step 16.5.1

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

$x = 8$

Step 16.5.2

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

$\frac{{(8)}^{2} - 9 \cdot 8 + 18}{4 {(8)}^{2} - 25} \leq 0$

Step 16.5.3

The left side $0.\overline{043290}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 16.6

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

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < \frac{5}{2}$ True

$\frac{5}{2} < x < 3$ False

$3 < x < 6$ True

$x > 6$ False

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < \frac{5}{2}$ True

$\frac{5}{2} < x < 3$ False

$3 < x < 6$ True

$x > 6$ False

Step 17

The solution consists of all of the true intervals.

$- \frac{5}{2} < x < \frac{5}{2}$ or $3 \leq x \leq 6$

Step 18

Convert the inequality to interval notation.

$(- \frac{5}{2}, \frac{5}{2}) \cup [3, 6]$

Step 19