Prove r divides b . . .?
Let r (an integer) be a nonzero solution of x^2 + ax + b =0 where a,b are integers
Prove r divides b
Not sure how to get to this solution...
- Anonymous8 years agoFavorite Answer
r solves x^2 + ax + b = 0
=> r*r + a*r = -b
=> (r+a)*r = -b
Since r,a are integers, -b is an integer multiple of r, and so is b, hence r divides b.
- 8 years ago
quadratic functions can be factorized, for example
x^2 + 3x +2 = 0, this can be factorized to
(x + 2)(x +1) = 0
and as we can see both 1 and 2 are solutions to this and both 1 and 2 divide 2,
so if we take x^2 + ax +b
you've already been given that r is a solution, so lets assume that some constant c (an integer) is also a solution, so we get
(x + r) ( x + c) = x^2 + ax +b = 0
as we can see from this r * c = b
then c = b/r
hence r divides b.
hope this helps.
- orometuaLv 48 years ago
y = (x-r)(x-s), so b=r*s and a=-(r+s). This shows that s is also an integer. Best of all, r | r*s so r|b.