Question

X^2 + 5 + 4 > 0

i have no idea how to solve this, please help<3

Answer 1

So the function is > 0 when x < -4 and when x > -1.

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.

Answer 2

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)