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

Lightning
Jul 24, 2018

#2**+1 **

5x^2 + 19x + 16 > 20 subtract 20 from both sides

5x^2 + 19x - 4 > 0 let's set this to 0

5x^2 + 19x - 4 = 0 factor, if we can

(5x -1 ) ( x + 4) = 0 set each factor to 0 and solve for x and we get

x =1/5 and x = -4

Note that we have three possible intervals that make the original inequality true

(-inf, -4) or ( - 4, 1/5) or ( 1/5, inf )

Normally, in a quadratic...if the middle interval makes the inequality untrue, the outside intervals will make it true

Testing a point in the middle interval - I'll pick x = 0 - and subbing it into the original inequality produces 16 > 20

So....the middle interval makes the inequality untrue....and the integers in this interval are -3, -2, -1 and 0 [ 4 values]

(Note that we can't include -4 since it is not part of the interval itself)

See the graph, here : https://www.desmos.com/calculator/dkmaryyfcp

CPhill
Jul 24, 2018