How do you solve #x^2+5x+4<=0#?

1 Answer
Jun 26, 2016

Answer:

#x in [-4, -1]#

Explanation:

#x^2+5x+4 = (x+4)(x+1)#

This is zero when #x=-4# or #x=-1#

The quadratic has a positive leading coefficient, so is is an upright parabola, crossing the #x# axis at #(-4, 0)# and #(-1, 0)#.

Hence any #x in [-4, -1]# satisfies the required inequality and any value of #x# outside that closed interval does not.

graph{x^2+5x+4 [-10, 10, -5, 5]}