Question

Find three consecutive positive odd integers such

that the square of the smallest exceeds twice the
largest by 7.

Answer 1

5, 7, 9

Explanation:

You want three consecutive positive odd integers such that the square of the smallest exceeds twice the largest by 7.

Setup

Let x represent the middle of the integers. Then x-2 is the smallest, and x+2 is the largest. The given relation is ...

(x -2)² -2(x +2) = 7

Solution

Eliminating parentheses gives ...

x² -4x +4 -2x -4 = 7

x² -6x -7 = 0 . . . . . . . . . subtract 7, simplify

(x -7)(x +1) = 0 . . . . . . . factor

Values of x that make the factors zero are solutions to this equation:

x -7 = 0 ⇒ x = 7

x +1 = 0 ⇒ x = -1

The positive odd numbers are 5, 7, 9.

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Check

5² -2(9) = 25 -18 = 7 . . . . as required