Originally Posted by

**topsquark** Well if the roots are to be integers then the discriminant has to be greater than or equal to 0. So:

(-a)^2 - 4*1*12 >= 0

a^2 >= 48

a >= sqrt(48) = 4*sqrt(3)

So two real roots are possible.

What are the roots of the equation in terms of a?

x = (a (+/-) sqrt(a^2 - 48))/2

If these two roots are to be consecutive integers then

x- = (a - sqrt(a^2 - 48))/2

x+ = (a + sqrt(a^2 - 48))/2

differ by 1.

So let (x+) = (x-) + 1

(a + sqrt(a^2 - 48))/2 = (a - sqrt(a^2 - 48))/2 + 1

a + sqrt(a^2 - 48) = a - sqrt(a^2 - 48) + 2

2*sqrt(a^2 - 48) = 2

sqrt(a^2 - 48) = 1

a^2 - 48 = 1

a^2 = 49

a = (+/-)7

Now, which one of these work? The easiest way is simply to solve the quadratics:

a = 7: x^2 - 7x + 12 = 0

Has solutions x = 3, 4. So a = 7 works.

a = -7: x^2 + 7x + 12 = 0

Has solutions x = -3, -4. So a = -7 also works.

-Dan