Question

If the zeroes of p(x)=x^2+ ax+ b are two consecutive integers then prove that a^2= 1+4b.

Answer 1

Answer:

Hello.

Hope this helps you.

The roots are n and n+1 for some integer n.

The sum of the roots is

2n+1 = -a

The product of the roots is

n(n+1) = b

Therefore

a² = (-a)² = (2n+1)²

= 4n² + 4n + 1

= 4n(n+1) + 1

= 1 + 4b