How do you solve #x^2 + 2x -3 <=0#?

1 Answer
May 7, 2015

First, solve the quadratic equation:
f(x) = x^2 + 2x - 3 = 0.
Since a + b + c = 0, the 2 real roots are: #1 and (c/a = -3)#
Try the test point x = 0 -> f(0) = -3 < 0. It is true, then the origin O is located in the right segment.
The interval [-3, 1] is the solution set of #f(x) <= 0.#
The 2 end (critical) points (-3) and (1) are included in the solution set.