X^2 + 5 + 4 > 0

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

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)

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.

i dont know nkw

step-by-step explanation:

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answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people