Solve the Inequality for x x^2>=5x+6

$x^{2} \geq 5 x + 6$

Step 1

Subtract $5 x$ from both sides of the inequality.

$x^{2} - 5 x \geq 6$

Step 2

Convert the inequality to an equation.

$x^{2} - 5 x = 6$

Step 3

Subtract $6$ from both sides of the equation.

$x^{2} - 5 x - 6 = 0$

Step 4

Factor $x^{2} - 5 x - 6$ 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 $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Write the factored form using these integers.

$(x - 6) (x + 1) = 0$

Step 5

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

$x - 6 = 0$

$x + 1 = 0$

Step 6

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 7

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

The final solution is all the values that make $(x - 6) (x + 1) = 0$ true.

$x = 6, - 1$

Step 9

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 6$

$x > 6$

Step 10

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{2} \geq 5 (- 4) + 6$

The left side $16$ is greater than the right side $- 14$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} \geq 5 (0) + 6$

The left side $0$ is less than the right side $6$ , which means that the given statement is false.

False

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

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

$x = 8$

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

${(8)}^{2} \geq 5 (8) + 6$

The left side $64$ is greater than the right side $46$ , which means that the given statement is always true.

True

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

$x < - 1$ True

$- 1 < x < 6$ False

$x > 6$ True

$x < - 1$ True

$- 1 < x < 6$ False

$x > 6$ True

Step 11

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 6$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1 or x \geq 6$

Interval Notation:

$(- \infty, - 1] \cup [6, \infty)$

Step 13

Solve the Inequality for x x^2>=5x+6

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