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?
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)
+129899 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
+373 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
