Solve the Inequality for x x^2-15x+36<=-2x

$x^{2} - 15 x + 36 \leq - 2 x$

Move all terms containing $x$ to the left side of the inequality.

Add $2 x$ to both sides of the inequality.

$x^{2} - 15 x + 36 + 2 x \leq 0$

Add $- 15 x$ and $2 x$ .

$x^{2} - 13 x + 36 \leq 0$

Convert the inequality to an equation.

$x^{2} - 13 x + 36 = 0$

Factor $x^{2} - 13 x + 36$ using the AC method.

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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Write the factored form using these integers.

$(x - 9) (x - 4) = 0$

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

$x - 9 = 0$

$x - 4 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

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

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x - 9) (x - 4) = 0$ true.

$x = 9, 4$

Use each root to create test intervals.

$x < 4$

$4 < x < 9$

$x > 9$

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 $x < 4$ to see if it makes the inequality true.

Choose a value on the interval $x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 15 \cdot 0 + 36 \leq - 2 \cdot 0$

The left side $36$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $4 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 9$ and see if this value makes the original inequality true.

$x = 6$

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

${(6)}^{2} - 15 \cdot 6 + 36 \leq - 2 \cdot 6$

The left side $- 18$ is less than the right side $- 12$ , which means that the given statement is always true.

True

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

${(12)}^{2} - 15 \cdot 12 + 36 \leq - 2 \cdot 12$

The left side $0$ is greater than the right side $- 24$ , which means that the given statement is false.

False

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

$x < 4$ False

$4 < x < 9$ True

$x > 9$ False

$x < 4$ False

$4 < x < 9$ True

$x > 9$ False

The solution consists of all of the true intervals.

$4 \leq x \leq 9$

The result can be shown in multiple forms.

Inequality Form:

$4 \leq x \leq 9$

Interval Notation:

$[4, 9]$

Solve the Inequality for x x^2-15x+36<=-2x

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