Question 651527
If the square of the larger of two consecutive integers is reduced by three
 times the smaller, the result is the sum of the integers.
 Find the integers
:
Let x = the larger integer
then
(x-1) = the smaller
:
x^2 - 3(x-1) = x + (x-1)
x^2 - 3x + 3 = 2x - 1
x^2 - 3x - 2x + 3 + 1 = 0
x^2 - 5x + 4 = 0
Factors to
(x-1)(x-4) = 0
two solutions
x = 1, then 0 = the smaller integer
and
x = 4, then 3 = the smaller integer
:
Both solutions seems to work, if you consider 0 an integer