Solve the Inequality for x (x+6)(x-7)(x^2+4)>=0

$(x + 6) (x - 7) (x^{2} + 4) \geq 0$

Step 1

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

$x + 6 = 0$

$x - 7 = 0$

$x^{2} + 4 = 0$

Step 2

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 4

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{4}$

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 i, - 2 i$

Step 5

The final solution is all the values that make $(x + 6) (x - 7) (x^{2} + 4) \geq 0$ true.

$x = - 6, 7, 2 i, - 2 i$

Step 6

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 7$

$x > 7$

Step 7

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$((- 8) + 6) ((- 8) - 7) ({(- 8)}^{2} + 4) \geq 0$

The left side $2040$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 6 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 6) ((0) - 7) ({(0)}^{2} + 4) \geq 0$

The left side $- 168$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 7$ to see if it makes the inequality true.

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

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

$((10) + 6) ((10) - 7) ({(10)}^{2} + 4) \geq 0$

The left side $4992$ 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.

$x < - 6$ True

$- 6 < x < 7$ False

$x > 7$ True

$x < - 6$ True

$- 6 < x < 7$ False

$x > 7$ True

Step 8

The solution consists of all of the true intervals.

$x \leq - 6$ or $x \geq 7$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 6 or x \geq 7$

Interval Notation:

$(- \infty, - 6] \cup [7, \infty)$

Step 10

Solve the Inequality for x (x+6)(x-7)(x^2+4)>=0

Related Questions

Support LCMGCF.com by disabling your adblocker.