Question

How do you solve `x² + 2x - 15 ≤ 0`?

Answer 1

The solution is `x in [-5, 3]`

Explanation:

The inequality is

`x^2+2x-15<=0`

`(x+5)(x-3)<=0`

Let `f(x)=(x+5)(x-3)`

Build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-5` `color(white)(aaaaaaaa)` `3` `color(white)(aaaaaa)` `+oo` ### ###

`color(white)(aaaa)` `x+5` `color(white)(aaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaaaa)` `+` ### ###

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaa)#`-` `color(white)(aaa)` `0` `color(white)(aaa)` `+`##

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+`##

Therefore,

`f(x)<=0` when `x in [-5, 3]`

graph{x^2+2x-15 [-36.52, 36.52, -18.26, 18.28]}