SOLUTION: Find three consecutive numbers, the sum of whose squares is 29 more than three times the square of the smallest.

Question 851069: Find three consecutive numbers, the sum of whose squares is 29 more than three times the square of the smallest.
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
Here's what we know from the problem:
A = B-1
B = C-1 (therefore C = A+2, B = A+1)
a^2 + b^2 + c^2 = 29 + 3a^2
.
Let's replace the values in the bottom equation with the equivalents we know.
a^2 + (a+1)^2 + (a+2)^2 = 29 + 3a^2
Square out the parentheses
a^2 + (a^2 + 2a + 1) + (a^2 + 4a + 4) = 29 + 3a^2
Add
3a^2 + 6a + 5 = 29 + 3a^2
Subtract 3a^2 from each side
6a + 5 = 29
Subtract 5 from each side
6a = 24
Divide each side by 6
a = 4
.
If a is 4, then b is 5, and c is 6
.
Let's plug it in "the sum of whose squares is 29 more than three times the square of the smallest."
4^2 + 5^2 +6^2 = 29 + (3*4^2)
16 + 25 + 36 = 29 + 48
41 + 36 = 77
77 = 77
Success!

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