For how many integer values of x is $5x^{2}+19x+16 > 20$ not satisfied?
Thx whoever answers this u just made my day!
+128474 5x^{2}+19x+16 > 20 subtract 20 from both sides
5x^2 + 19x - 4 > 0 (1)
Let's solve this.....and then we can find the interval that works
5x^2 + 19x - 4 = 0 factor
(5x - 1) ( x + 4) = 0
Setting each factor to 0 and solving for x we get that
x = 1/5 or x = 4
We have three possible intervals that might solve (1)
(-infinity, -4) or (-4, 1/5) or ( 1/5, infinity)
Usually, either the middle interval works or the two outside intervals work
If we pick a point in the middle interval - I'll choose 0 - note that this value makes (1) false
So....the intervals that solve this are
(-infinity, 4) U ( 1/5 , infinity)
