Question

How do you solve the inequality `x^2+2x-3>=0` and write your answer in interval notation?

Answer 1

The solution is `x in (-oo, -3] uu [1, +oo)`

Explanation:

Let's factorise the inequality

`x^2+2x-3>=0`

`(x+3)(x-1)>=0`

Let `f(x)=(x+3)(x-1)`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaaaaaaa)` `1` `color(white)(aaaaa)` `+oo`

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

`color(white)(aaaa)` `x-1` `color(white)(aaaaaa)` `-` `color(white)(aaa)` ###color(white)(aaaaa)#`-` `color(white)(aaa)` `0` `color(white)(aa)` `+`

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

Therefore,

`f(x)>=0` when `x in (-oo, -3] uu [1, +oo)`

graph{x^2+2x-3 [-10, 10, -5, 5]}