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

 

Thx whoever answers this u just made my day!

 Feb 23, 2019
 #1
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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)

 

 

cool  cool cool

 Feb 23, 2019

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