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

2 Answers
Feb 21, 2016

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]#

Feb 22, 2016

Answer:

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.