Question

14. Find three consecutive odd integers such that the product of the first and the third is 13 less

than ten times the second integer.

Answer 1

{-1, 1, 3} or {7, 9, 11}

Explanation:

Let x represent the middle integer. Then the first is (x-2) and the third is (x+2). The given relation is ...

(x -2)(x +2) = 10x -13

x^2 -4 = 10x -13 . . . . . . . eliminate parentheses

x^2 -10x = -9 . . . . . . . . . . add 4-10x

x^2 -10x +25 = 16 . . . . . add 25 to complete the square

(x -5)^2 = 16 . . . . . . . . show the left side as a square

x -5 = ±4 . . . . . . . . . . take the square root

x = 5 ± 4 = {1, 9}

There are two solutions:

{-1, 1, 3} or {7, 9, 11}