Question

How do you solve `(x + 2) ( 4x - 3) \leq 0`?

Answer 1

The solution is `x in [-2,3/4]`

Explanation:

Let `f(x)=(x+2)(4x-3)`

Let 's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaaaa)` `3/4` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `4x-3` `color(white)(aaaaaa)` `-` `color(white)(aa)` ###color(white)(aaaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aa)` `0` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

Therefore,

`f(x)<=0`, when `x in [-2,3/4]`

graph{(x+2)(4x-3) [-11.26, 11.24, -7.88, 3.37]}