How do you solve #x^2+ 4x + 3 >= 0#?
1 Answer
May 22, 2015
y = x^2 + 4x +3 >= 0
Since (a - b + c) = 0, one real root is (-1) and the other is (-c/a = -3).
The parabola opens upward (a = 1 > 0). Between the 2 real root, y is negative. Then, y is positive (> 0) out side the interval (-3, -1).
The solution set is the half closed intervals: (-infinity, -3] and [-1, +infinity). The 2 end points, or critical points, are included in the solution set.
You may also solve the inequality by the number line and test point method. Substitute x = 0 of the origin into y, we get: 3 > 0. True.
Then O is on the true ray. By symmetry, the other ray also belongs to the solution set.
===============|-3-----------|-1==|0=============
