Solve the Inequality for x x^2+5x-24>=0

$x^{2} + 5 x - 24 \geq 0$

Convert the inequality to an 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 $5$ .

$- 3, 8$

Write the factored form using these integers.

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

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

$x - 3 = 0$

$x + 8 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

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

$x = 3, - 8$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 3$

$x > 3$

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 $x < - 8$ 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.

${(- 10)}^{2} + 5 (- 10) - 24 \geq 0$

The left side $26$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $- 24$ is less than the right side $0$ , 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.

${(6)}^{2} + 5 (6) - 24 \geq 0$

The left side $42$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 8$ True

$- 8 < x < 3$ False

$x > 3$ True

$x < - 8$ True

$- 8 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve the Inequality for x x^2+5x-24>=0

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