# How do you solve the inequality x^2+2x>=24?

May 9, 2017

By trial and error. Your answer is with the exception of -6<x<4, all x values provide solution.

#### Explanation:

x must be a real number and let me try +3 provide the solution. There are two numbers provide the solution (one is positive and the other one is negative):

${3}^{2} + 2 \cdot 3 = 15$ therefore it does not provide solution.

Now let me try +4

${4}^{2} + 2 \cdot 4 = 24$ Yes. This is one answer. Therefore x must be greater than or equal to +4.

On the negative side, let me try -3 first:

${\left(- 3\right)}^{2} - 2 \cdot 3 = 2$ does not satisfy the requested condition.

Try -4 now:

${\left(- 4\right)}^{2} - 2 \cdot 4 = 8$ does not satisfy the requested condition.

Try -5 now:

${\left(- 5\right)}^{2} - 2 \cdot 5 = 15$ does not satisfy the requested condition.

Try -6 now:

${\left(- 6\right)}^{2} - 2 \cdot 6 = 24$ Yes it satisfies.

Any number less than -6 (including -6) will satisfy the given.

For instance (-8):

${\left(- 8\right)}^{2} - 2 \cdot 8 = 48$ which is greater than 24.

Now your solution is x must be greater than or equal to 4 or x must be less than or equal to -6.

May 9, 2017

The solution is $x \in \left(- \infty , - 6\right] \cup \left[4 , + \infty\right)$

#### Explanation:

We solve this equation with a sign chart

Let's rearrange and factorise the inequality

${x}^{2} + 2 x \ge 24$

${x}^{2} + 2 x - 24 \ge 0$

$\left(x + 6\right) \left(x - 4\right) \ge 0$

Let $f \left(x\right) = \left(x + 6\right) \left(x - 4\right)$

Now, we construct the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 6$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$, when $x \in \left(- \infty , - 6\right] \cup \left[4 , + \infty\right)$