Solve the Inequality for x x^2-6x<-9

$x^{2} - 6 x < - 9$

Convert the inequality to an equation.

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

Add $9$ to both sides of the equation.

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

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Use each root to create test intervals.

$x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = 0$

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

${(0)}^{2} - 6 \cdot 0 < - 9$

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

False

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

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

$x = 6$

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

${(6)}^{2} - 6 \cdot 6 < - 9$

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

False

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

$x < 3$ False

$x > 3$ False

$x < 3$ False

$x > 3$ False

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

No solution