Expert: Socrates - 5/13/2006

Question
Please help in explaning how to solve the following problem:
Find three consecutive integers such that the sum of their squares is 77?  Thank you

I am also stuck on how to solve; t^2 + 1 = 13/6t.

Please help.  thank you

Answer
We can write three consecutive integers as n-1 , n and n+1.
The sum of their squares must be 77 , so
(n-1)^2  + n^2 + (n+1)^2 = 77

n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 77

3n^2 + 2 = 77

3n^2 = 75

n^2 = 25

n = 5


substitute 5 for n in the expressions

n-1 , n , n+1

and get 4 , 5 , 6

The answer is 4,5,6

To solve t^2 + 1 = 13/6t , multiply both sides by 6t to   get rid of the fraction and get

6t^3 + 6t = 13

so

6t^3 + 6t - 13 = 0

to find rational solutions , try all factors of 13 divided by all factors of 6 , this gives 1 , 1/6 , 1/3 , 1/2 , -1 , -1/6 , -1/3 , -1/2 , 13 , 13/6 , 13/3 , 13/2 , -13 , -13/6 , -13/3 , -13/2 . Unfortunately, you must try these  to find a solution. Since it appears that none of these work, the equation has no rational roots. The best you can do now is to look for an approximation. Since this is rather involved, I suspect that the equation you gave me is incorrect. In any event, one solution of the equation you gave is approximately 1.04.