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Graphing Funtions

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

Jul 24, 2018

#1
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Thank you Chris for getting back to me so fast!

Jul 24, 2018
#2
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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

Jul 24, 2018
#3
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No prob...maybe they should call Me "Lightning,"  huh ??

Jul 24, 2018
#4
+1206
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Lol! Maybe i should! BTW, -4 is included! Thanks!

Lightning  Jul 24, 2018
#5
+102383
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Yep...you are correct...I forgot that we  need to  find the integer values that make the inequality untrue...-4 should be included!!!

CPhill  Jul 24, 2018