How do you solve #x^2+6x+5<=0#?

1 Answer
May 22, 2015

#f(x) = x^2 + 6x + 5 <= 0#
Since (a - b + c) = 0, one real root is (-1) and the other is (-c/a = -5). There are 2 common methods to solve this inequalities:
1. Using number line and test point method.
Plot the 2 real roots (-5) and (-1) on a number line. They divide the line into one segment (-5, -1) and two rays. Always try the origin O as test point. x = 0 --> f(x) = 5 < 0. Not true, then O isn't located on the solution set. The solution set is the closed interval [-5, -1]. The 2 end points are included in the solution set.
2. Using The Theorem:
"Between the 2 real roots, f(x) has the opposite sign with a".
In this example, between the 2 real roots (-5) and (-1), f(x) is negative (< 0), as opposite in sign to a = 1 (> 0). Then, the solution set is the closed interval [-5, -1]

<-----------------|-5==========|-1-------|0------------------->