Question

What is the solution set for `-x^2 + 2x > -3`?

Answer 1

`x in (-1,3)`

Explanation:

Start by getting all the terms on one side of the inequality. You can do that by adding `3` to both sides

`-x^2 + 2x + 3 > - color(red)(cancel(color(black)(3))) + color(red)(cancel(color(black)(3)))`

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

Next, make the quadratic equal to zero in order to find its roots. This will help you factor it. Use the quadratic formula to calculate `x_(1,2)`.

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

`x_(1,2) = (-2 +- sqrt(2^2 - 4 * (-1) * (3)))/(2 * (-1))`

`x_(1,2) = (-2 +- sqrt(16))/((-2))`

`x_(1,2) = (-2 +- 4)/((-2)) = {(x_1 = (-2-4)/((-2)) =3), (x_2 = (-2 + 4)/((-2)) = -1) :}`

This means that you can rewrite the quadratic as

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

Your inequality will be equivalent to

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

In order for this inequality to be true, you need one of the two terms to be positive and the other negative, or vice versa.

Your first two conditions will be

`x-3 > 0 implies x > 3`

and

`x + 1 < 0 implies x < -1`

Since you can't have values of `x` that are both greater than `3` and smaller than `(-1)`, this possibility is eliminated.

The other conditions wll be

`x - 3 < 0 implies x < 3`

and

`x + 1 > 0 implies x > -1`

This time, these two intervals will produce a valid solution set. For any value of `x` that is greater than `(-1)` and smaller than `3`, this product

`(x-3) * (x+1) <0`

which means that

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

The solution set for this inequality will thus be `x in (-1,3)`.