Answer:
Step-by-step explanation:
Solve x^2 + 5x + 4 = 0 first. This quadratic can be factored into
(x + 1)(x + 4) = 0, whose roots are x = -1 and x = -4. Set up three intervals on the number line based on x: {-4, -1}: (-infinity, -4), (-4, -1), (-1, infinity).
We must now determine whether x^2 + 5 + 4 > 0 is true on each interval. We choose a test number from each interval: from (-infinity, -4) choose the test number -5; from (-4, -1) choose -3, and from (-1, infinity) choose 5.
Case 1: x = -5: (-5)^2 + 5(-5) + 4 > 0 is true. Thus, one part of the solution set is (-infinity, -4).
Case 2: x = -3: (-3)^2 - 25 + 4 > 0) is false. Thus, no solutions lie in the interval (-4, -1).
Case 3: x = 5: (5)^2 + 5(5) + 4 is true. Thus, (-1, infinity) is also part of the solution set.
This two-part solution set can be written as
(-infinity, -4) ∪ (-1, infinity)
Answer:
So the function is > 0 when x < -4 and when x > -1.
Step-by-step explanation:
First find the critical values of x:
x^2 + 5x + 4 = 0
(x + 1)(x + 4) = 0
x = -4, -1.
Make a table:
x < -4 -4<x<-1 x > -1
x + 1 < 0 <0 >0
x + 4 < 0 >0 >0
(x+4)(x+1) >0 <0 > 0
So the function is > 0 when x < -4 and x > -1.
