Solve the Inequality for x x^2-3x-10>=0

$x^{2} - 3 x - 10 \geq 0$

Convert the inequality to an equation.

$x^{2} - 3 x - 10 = 0$

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x - 5) (x + 2) = 0$

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

$x - 5 = 0$

$x + 2 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 5$

$x > 5$

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 $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{2} - 3 \cdot - 4 - 10 \geq 0$

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 3 \cdot 0 - 10 \geq 0$

The left side $- 10$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

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

${(8)}^{2} - 3 \cdot 8 - 10 \geq 0$

The left side $30$ 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.

$x < - 2$ True

$- 2 < x < 5$ False

$x > 5$ True

$x < - 2$ True

$- 2 < x < 5$ False

$x > 5$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 5$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq 5$

Interval Notation:

$(- \infty, - 2] \cup [5, \infty)$

Solve the Inequality for x x^2-3x-10>=0

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