Solve the Inequality for x x^2+x-20>=0

$x^{2} + x - 20 \geq 0$

Convert the inequality to an equation.

$x^{2} + x - 20 = 0$

Factor $x^{2} + x - 20$ 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 $- 20$ and whose sum is $1$ .

$- 4, 5$

Write the factored form using these integers.

$(x - 4) (x + 5) = 0$

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

$x - 4 = 0$

$x + 5 = 0$

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.

$x = 4$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 4) (x + 5) = 0$ true.

$x = 4, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 4$

$x > 4$

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 $x < - 5$ 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.

${(- 8)}^{2} - 8 - 20 \geq 0$

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

True

Test a value on the interval $- 5 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 0 - 20 \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

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.

$x = 6$

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

${(6)}^{2} + 6 - 20 \geq 0$

The left side $22$ 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.

$x < - 5$ True

$- 5 < x < 4$ False

$x > 4$ True

$x < - 5$ True

$- 5 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$x \leq - 5$ or $x \geq 4$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 5 or x \geq 4$

Interval Notation:

$(- \infty, - 5] \cup [4, \infty)$

Solve the Inequality for x x^2+x-20>=0

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