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...

3 Answers

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  • Anonymous
    7 years ago
    Favorite 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.

  • 7 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.

  • 7 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.

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