Problema Solution
Find three consecutive integers, the sum of whose squares is 65 more than three times the square of the smallest.
Answer provided by our tutors
X^2+(x+1)^2+(x+2)^2 = 3x^2 +65
Or, 6x+5 =65
So x = 10
Numbers are 10, 11, 12
Find three consecutive integers, the sum of whose squares is 65 more than three times the square of the smallest.
X^2+(x+1)^2+(x+2)^2 = 3x^2 +65
Or, 6x+5 =65
So x = 10
Numbers are 10, 11, 12