Solve the Inequality for x x^2>=5x+24

$x^{2} \geq 5 x + 24$

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

$x^{2} - 5 x \geq 24$

Convert the inequality to an equation.

$x^{2} - 5 x = 24$

Subtract $24$ from both sides of the equation.

$x^{2} - 5 x - 24 = 0$

Factor $x^{2} - 5 x - 24$ 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 $- 24$ and whose sum is $- 5$ .

$- 8, 3$

Write the factored form using these integers.

$(x - 8) (x + 3) = 0$

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

$x - 8 = 0$

$x + 3 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 8) (x + 3) = 0$ true.

$x = 8, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 8$

$x > 8$

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.

$x = - 6$

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

${(- 6)}^{2} \geq 5 (- 6) + 24$

The left side $36$ is greater than the right side $- 6$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} \geq 5 (0) + 24$

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

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

${(10)}^{2} \geq 5 (10) + 24$

The left side $100$ is greater than the right side $74$ , which means that the given statement is always true.

True

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

$x < - 3$ True

$- 3 < x < 8$ False

$x > 8$ True

$x < - 3$ True

$- 3 < x < 8$ False

$x > 8$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq 8$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x \geq 8$

Interval Notation:

$(- \infty, - 3] \cup [8, \infty)$

Solve the Inequality for x x^2>=5x+24

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