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Three consecutive positive odd integers a,b and c satisfy a^2-b^2=\(32\) and c^{2}-b^{2}>0. What is the value of c^{2}-b{^2}?

 Apr 18, 2022
 #1
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hey hope you have a good day

 Apr 18, 2022
 #2
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are you the person who posted this question

 Apr 18, 2022
 #3
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a^2 - b^2  =  32     

b^2 - c^2  >  0

 

If a^2  - b^2  =32, then a >b

And b > c

 

Let a =  2n + 1    and   b  =2n - 1     and   c =  2n - 3

 

So

 

(2n + 1) ^2  -  (2n  - 1)^2  = 32       simplify

 

4n^2  + 4n + 1   - 4n^2 + 4n  - 1   = 32

 

8n = 32

 

n = 4

 

c = 2(4)  - 3   = 5

b = 2(4) - 1   =  7

a = 2(4) + 1  =  9

 

So

 

b^2   - c^2   =  

 

7^2  -  5^2   =

 

49   -  25   = 

 

24

 

 

cool cool cool

 Apr 18, 2022

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