Question

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

Answer 1

`x<=(-7)/2 or x>=0`

Explanation:

`x(2x+7)>=0`

Distribute

`(x)(2x)+(x)(7)>=0`

`2x^2 + 7x >=0`

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

`2x^2 + 7x =0`

We need to factorize this again

#x(2x+7)=0

Set factors equal to `0`

`x=0 or 2x+7=0`

`x=0 or 2x=0-7`

`x=0 or 2x=-7`

`x=0 or x=(-7)/2`

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

`x<=(-7)/2` (This works in original inequality)

`(-7)/2<=x<=0` (This doesn't work)

`x>=0` (This works)

Thus,

The answer is:

`x<=(-7)/2 or x>=0`

Answer 2

Use the Sign Chart Method

):

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,-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(2*-4+7)>=0->"negative times negative is positive, this works"`

`-2(2*-2+7)>=0->"negative times positive is negative, does not work"`

`2(2*2+7)>=0->"positive times positive is positive, this works"`

Write in interval notation:

`(-∞,-7/2]U[0,∞)`

(Use brackets as the answer can be equal to)