Question

How do you solve `x^2+2x-3>=0`?

Answer 1

`x>=1, x<=-3`

Explanation:

The key to questions like this one is to solve it in the same way that you solve an equality (so one with an `=` sign), with two caveats:

• that if you multiply or divide by a negative number, you flip the greater than sign, and
• we'll be setting zones to see what is valid and what isn't.

So starting with the original:

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

we can factor this into:

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

and solve for x:

`x>=-3` and `x>=1`

Ok, so we know that we have 3 zones: `x>=1`, `x<=-3`, and the region between 1 and `-3`. Let's test `x=-2` to see if the middle region is valid:

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

`(-2)^2+2(-2)-3>=0`

`4-4-3>=0`

`-3>=0` - nope.

And let's test `x=-4` to test the left-most region:

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

`(-4)^2+2(-4)-3>=0`

`16-8-3>=0`

`5>=0` - yes.

And we can test the right-most region with `x=2`:

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

`2^2+2(2)-3>=0`

`4+4-3>=0`

`5>=0` - yes.

So the answer in notation is `x>=1, x<=-3`. The graph is below (on a number line, this will be solid lines on the x-axis in the shaded area starting from a solid point at x=1 and going right and another from x=-3 and going left.)

graph{x^2+2x-3>=0 [-10, 10, -1, 1]}