Precalculus Examples

$6 x - 9 - x^{2} > 0$

Step 1

Convert the inequality to an equation.

$6 x - 9 - x^{2} = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $- 1$ out of $6 x - 9 - x^{2}$ .

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Step 2.1.1

Reorder the expression.

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Step 2.1.1.1

Move $- 9$ .

$6 x - x^{2} - 9 = 0$

Step 2.1.1.2

Reorder $6 x$ and $- x^{2}$ .

$- x^{2} + 6 x - 9 = 0$

Step 2.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 6 x - 9 = 0$

Step 2.1.3

Factor $- 1$ out of $6 x$ .

$- (x^{2}) - (- 6 x) - 9 = 0$

Step 2.1.4

Rewrite $- 9$ as $- 1 (9)$ .

$- (x^{2}) - (- 6 x) - 1 \cdot 9 = 0$

Step 2.1.5

Factor $- 1$ out of $- (x^{2}) - (- 6 x)$ .

$- (x^{2} - 6 x) - 1 \cdot 9 = 0$

Step 2.1.6

Factor $- 1$ out of $- (x^{2} - 6 x) - 1 (9)$ .

$- (x^{2} - 6 x + 9) = 0$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

Rewrite $9$ as $3^{2}$ .

$- (x^{2} - 6 x + 3^{2}) = 0$

Step 2.2.2

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

$6 x = 2 \cdot x \cdot 3$

Step 2.2.3

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 3 + 3^{2}) = 0$

Step 2.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$- {(x - 3)}^{2} = 0$

Step 3

Divide each term in $- {(x - 3)}^{2} = 0$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- {(x - 3)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 3)}^{2}}{- 1} = \frac{0}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{{(x - 3)}^{2}}{1} = \frac{0}{- 1}$

Step 3.2.2

Divide ${(x - 3)}^{2}$ by $1$ .

${(x - 3)}^{2} = \frac{0}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $0$ by $- 1$ .

${(x - 3)}^{2} = 0$

Step 4

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

$x - 3 = 0$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Use each root to create test intervals.

$x < 3$

$x > 3$

Step 7

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 7.1

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

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Step 7.1.1

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

$x = 0$

Step 7.1.2

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

$6 (0) - 9 - {(0)}^{2} > 0$

Step 7.1.3

The left side $- 9$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 7.2

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

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Step 7.2.1

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

$x = 6$

Step 7.2.2

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

$6 (6) - 9 - {(6)}^{2} > 0$

Step 7.2.3

The left side $- 9$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 7.3

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

$x < 3$ False

$x > 3$ False

$x < 3$ False

$x > 3$ False

Step 8

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

No solution