Question 165701
Find the three consecutive odd integers such that the product of the first and third integers is 4 less than the square of the second integer
:
The three odd integers: x, (x+2), (x+4)
:
1st*3rd = 2nd^2 less 4
x*(x+4) = (x+2)^2 - 4
:
x^2 + 4x = x^2 + 4x +4 - 4
:
x^2 + 4x = x^2 + 4x; no kidding!
:
No unique solution, any value for x will make the equation happy