# How do you solve x(2x+7) >=0?

Mar 4, 2018

$x \le \frac{- 7}{2} \mathmr{and} x \ge 0$

#### Explanation:

$x \left(2 x + 7\right) \ge 0$

Distribute

$\left(x\right) \left(2 x\right) + \left(x\right) \left(7\right) \ge 0$

$2 {x}^{2} + 7 x \ge 0$

Now we need to find the critical points of the inequality!

$2 {x}^{2} + 7 x = 0$

We need to factorize this again

x(2x+7)=0

Set factors equal to $0$

$x = 0 \mathmr{and} 2 x + 7 = 0$

$x = 0 \mathmr{and} 2 x = 0 - 7$

$x = 0 \mathmr{and} 2 x = - 7$

$x = 0 \mathmr{and} x = \frac{- 7}{2}$

Now we need to check the intervals between the critical points!

$x \le \frac{- 7}{2}$ (This works in original inequality)

$\frac{- 7}{2} \le x \le 0$ (This doesn't work)

$x \ge 0$ (This works)

Thus,

$x \le \frac{- 7}{2} \mathmr{and} x \ge 0$

Mar 4, 2018

Use the Sign Chart Method

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#### Explanation:

Factor (make sure it is equal to zero) and Solve; this is already factored and equal to zero:
Pretend the inequality sign is an equal sign:

$x = 0 , - \frac{7}{2}$

Plot these on a number line: The two points create $3$ sections. Plug in any point into the inequality from each section and check the sign:

$- 4 \left(2 \cdot - 4 + 7\right) \ge 0 \to \text{negative times negative is positive, this works}$

$- 2 \left(2 \cdot - 2 + 7\right) \ge 0 \to \text{negative times positive is negative, does not work}$

$2 \left(2 \cdot 2 + 7\right) \ge 0 \to \text{positive times positive is positive, this works}$

Write in interval notation:

(-∞,-7/2]U[0,∞)#

(Use brackets as the answer can be equal to)