Solve for x x^2-3x-4>0

$x^{2} - 3 x - 4 > 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 3 x - 4 = 0$

Step 2

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

Step 3

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

$x - 4 = 0$

$x + 1 = 0$

Step 4

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.

$x = 4$

Step 5

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 6

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 7

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 4$

$x > 4$

Step 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 $x < - 1$ 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.

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

The left side $24$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $- 4$ is not greater 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.

${(6)}^{2} - 3 \cdot 6 - 4 > 0$

The left side $14$ 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.

$x < - 1$ True

$- 1 < x < 4$ False

$x > 4$ True

$x < - 1$ True

$- 1 < x < 4$ False

$x > 4$ True

Step 9

The solution consists of all of the true intervals.

$x < - 1$ or $x > 4$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x > 4$

Interval Notation:

$(- \infty, - 1) \cup (4, \infty)$

Step 11

Solve for x x^2-3x-4>0

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