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}?
+128475 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
