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For how many integer values of x is 5x^2 + 19x + 16 > 20 - 13x not satisfied?

 Apr 8, 2021
 #1
avatar+320 
+1

Hey there, Guest!

 

We can go minus both sides of the expression by 20 and obtain the following,

 

\(5x²+19x−4>05x²+19x−4>0\)

 

\((5x−1)(x+4)>0(5x−1)(x+4)>0\)

 

For the values that are not satisfied, we can take all the values between 1/5 and -4.

 

Values of x that are not satisfied =\([x:−4

which means that the answer to our problem is

\(x=[0,−1,−2,−3].x=[0,−1,−2,−3].\)

When counted, the number of elements, n(x), will give us 4.

And there we have our answer.

 

Hope this helped! :)

( ゚д゚)つ Bye

 Apr 8, 2021
 #2
avatar+121004 
+1

5x^2  +  19x  + 16   > 20  - 13x            rearrange  as

 

5x^2  + 32x  - 4   >    0

 

See  the  graph here   : https://www.desmos.com/calculator/bc6p7e68v8

 

This  will be true  except  on  this  approx interval   ( -6.523, .123)

 

So   7  integer  values

 

 

cool cool cool

 Apr 8, 2021

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