Precalculus Examples

$81 x^{2} - 18 x + 1 < 0$

Step 1

Convert the inequality to an equation.

$81 x^{2} - 18 x + 1 = 0$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $81 x^{2}$ as ${(9 x)}^{2}$ .

${(9 x)}^{2} - 18 x + 1 = 0$

Step 2.2

Rewrite $1$ as $1^{2}$ .

${(9 x)}^{2} - 18 x + 1^{2} = 0$

Step 2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 x = 2 \cdot (9 x) \cdot 1$

Step 2.4

Rewrite the polynomial.

${(9 x)}^{2} - 2 \cdot (9 x) \cdot 1 + 1^{2} = 0$

Step 2.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 9 x$ and $b = 1$ .

${(9 x - 1)}^{2} = 0$

Step 3

Set the $9 x - 1$ equal to $0$ .

$9 x - 1 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $1$ to both sides of the equation.

$9 x = 1$

Step 4.2

Divide each term in $9 x = 1$ by $9$ and simplify.

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Step 4.2.1

Divide each term in $9 x = 1$ by $9$ .

$\frac{9 x}{9} = \frac{1}{9}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $9$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{1}{9}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{9}$

Step 5

Use each root to create test intervals.

$x < \frac{1}{9}$

$x > \frac{1}{9}$

Step 6

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 6.1

Test a value on the interval $x < \frac{1}{9}$ to see if it makes the inequality true.

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Step 6.1.1

Choose a value on the interval $x < \frac{1}{9}$ and see if this value makes the original inequality true.

$x = 0$

Step 6.1.2

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

$81 {(0)}^{2} - 18 \cdot 0 + 1 < 0$

Step 6.1.3

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

False

Step 6.2

Test a value on the interval $x > \frac{1}{9}$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $x > \frac{1}{9}$ and see if this value makes the original inequality true.

$x = 3$

Step 6.2.2

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

$81 {(3)}^{2} - 18 \cdot 3 + 1 < 0$

Step 6.2.3

The left side $676$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 6.3

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

$x < \frac{1}{9}$ False

$x > \frac{1}{9}$ False

$x < \frac{1}{9}$ False

$x > \frac{1}{9}$ False

Step 7

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution