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Three consecutive positive odd integers a, b and c satisfy b^2 - a^2 = 348 and c^2 - b^2 > 0. What is the value of c^2 - b^2?

 Jul 14, 2021
 #1
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0

Let's assume A is the smallest. Then b would be A + 2, and C would be A + 4. Sub those values in. Can you take it from here? 

 

 

(I also hope i'm correct)

 Jul 14, 2021
 #2
avatar+121008 
+1

No such consecutive postivie odds   exist

 

Let  the  integers  be       2n - 1  ,  2n + 1 and 2n + 3

 

So

 

(2n + 1)^2  - (2n - 1)^2  =  348

 

4n^2  + 4n + 1  - 4n^2 + 4n - 1  = 348

 

8n =  348

 

n = 348/8   =  87/2

 

2 (87/2)  - 1 =      86

2n + 1  =  88

2n + 3  =  90

 

Note

 

88^2  - 86^2   =  348

 

And

 

90^2  -88^2  >   0

 

cool cool cool

 Jul 14, 2021
 #3
avatar+320 
+2

Hey there, Guest!

 

Alright, so let's write the individual worths of these variables:

 

a = a

b = a + 2

c = a + 4

 

So...

 

b^2 - a^2 = 348

 

(b+a)(b-a) = 348

 

(a+2+a)(a+2-a) = 348

 

Therefore a = 2.

 

Let's insert that into both of the equations for b and c.

 

b =a + 2

b = 2 + 2

b = 3

 

c = a +4

c = 2 + 4

c = 6

 

And here we are, back at the original equation, except now we know that:

 

a = 2

b = 3

c = 6

 

Let's insert that information into the question:

 

c^2 - b^2

6^2 - 3^2

36 - 9

= 27

 

Therefore the answer is 27.

 

Hope this helped! :)

( ゚д゚)つ Bye

 Jul 14, 2021

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