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# Please Help

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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

### 3+0 Answers

#1
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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

May 1, 2021
#2
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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
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TYSM! YOUR ALL CORRECT!

May 1, 2021