Question

How do you solve the inequality `x^2 + 2x - 3 <=0`?

Answer 1

First calculate the zeros of the polynom:

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

`x=(-2+-sqrt(16))/2`

`x=(-2+-4)/2`

`x=-3 or x=1`

So the inequality can be simplified into:

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

This will happen when one of the factors is positive and the other negative, so the solution will be:

`x in [-3, 1]`

Answer 2

Closed Interval [-3, 1]

Explanation:

`f(x) = x^2 + 2x - 3 <= 0`
First , solve f(x) = 0.
Since a + b + c = 0, use shortcut. One real root is 1 and the other is c/a = -3.
Use the algebraic method to solve f(x) <= 0. Between the 2 real roots (x-intercepts), f(x) < 0. (The parabola graph is below the x-axis because a > 0)).
Solution set: closed interval [-3, 1].
The 2 end points ( -3 and 1) are included in the solution set.