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What is the value of B if -x^2+bx-5<0 only when x\in (-\infty, 1)\cup(5,\infty)? I know that this is a repost but the other question did not have an answer. I have been stuck for a long time. Help Please

 Aug 9, 2019
 #1
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+1

\( x\in (-\infty, 1)\cup(5,\infty) \)

 

-x^2 + bx - 5 <  0    (1)      multiply though by  - 1   and  change the  direction of the  inequalitiy sign

 

x^2 - bx  + 5  >   0        set = to  0   

 

x^2 - bx + 5   =  0

 

We need to  have these  factors

 

(x - 1)  ( x - 5)     expanding we  get that

 

x^2  - 6x  +  5   =  0    so....this implies that we actually need

 

x^2 - 6x + 5   >  0        (2)

 

(x - 1) ( x - 5)  >  0

 

Note that  when    x =  ( -inf, 1)     both linear factors will be  negative.....so....their  product will be  positive

And  when x = ( 5, inf)   the same will occur

 

So  multiplying (2)  back  through by  -1   will produce  (1)   and we have that

 

-x^2 + 6x - 5 <  0 ....so......b  = 6

 

Here's a graph to show that this function is < 0   on  (-inf, 1)   and (5, inf)

 

https://www.desmos.com/calculator/uwy0jmousq

 

 

cool cool cool

 Aug 9, 2019
 #2
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+1

OMG THANK YOU SO MUCH. Thx. 

 Aug 9, 2019
 #3
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+1

Also I have never even used a website like this before.

Guest Aug 9, 2019
 #4
avatar+106515 
0

Well....welcome aboard!!!

 

Note....you could register and  become a member.....at no charge  !!!!

 

 

cool cool cool

CPhill  Aug 9, 2019

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