Question

Consider three consecutive positive integers, such that the sum of the

squares of the two larger integers is 5 more than 40 times the smaller
one. Find the smaller integer.

Answer 1

17

Explanation:

Let x represent the smaller integer. Then we have ...

(x +1)² +(x +2)² = 40x +5

2x² +6x +5 = 40x +5

x² -17x = 0 . . . . . subtract (40x+5), divide by 2

x(x -17) = 0 . . . . . factor

The solution of interest is x = 17.

The smaller integer is 17.