For how many integer values of x is $5x^{2}+19x+16 > 20$ not satisfied?

Thx whoever answers this u just made my day!

Guest Feb 23, 2019

#1**+1 **

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)

CPhill Feb 23, 2019