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frutty [35]
3 years ago
9

The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?

Mathematics
1 answer:
elixir [45]3 years ago
3 0
Let's say, the first number is "n"

well, the consecutive number of "n" is "n + 1"and the consecutive number of "n+1" is "n + 1" +1 or n + 2

so, the numbers are (n), (n+1) and (n+2), whatever "n" is

now, the sum of their squares is 116

\bf (n)^2+(n+1)^2+(n+2)^2=116

expand the binomials by FOIL or binomial theorem, and then simplify
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