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# help plz

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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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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
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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   Jul 14, 2021
#3
+5

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