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Find the greatest integer value of $b$ for which the expression $\frac{9x^3+4x^2+11x+7}{x^2+bx+8}$ has a domain of all real numbers.

 May 1, 2021
edited by WillBillDillPickle  May 1, 2021
 #1
avatar+121047 
+1

The denomiantor cannot  =  0 

 

So...we can guarantee  no  reals  will make the  denominator =  0   if we solve this :

 

b^2  - 4 (8)(1)  <  0

 

b^2  - 32  <  0

 

b^2  <  32        take  the positive root

 

b  < sqrt 32

 

b < 5.6

 

So  5  is  the  greatest  integer

 

 

cool cool cool

 May 1, 2021
 #2
avatar+795 
+1

Find the greatest integer value of b for which the expression (9x^3+4x^2+11x+7)/(x^2+bx+8) has a domain of all real numbers.

 

Domain is all real numbers iff x^2+bx+8 ≠ 0.

 

Let x^2 + bx + 8 = 0 and then we will find a discriminant that will result in no real solutions if x^2+bx+8 = 0 because the denominator ≠ 0.

 

A equation is unreal iff b^2 - 4ac < 0

So b^2 - 32 < 0

b^2 < 32

floor(sqrt(32)) = 5

 

5 is the answer I think

 #3
avatar+309 
0

TYSM! YOUR ALL CORRECT! smiley

 May 1, 2021

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