Find the condition that one root of the quadratic equation may be 1 more than the other.

Let p, np be the roots of the given QE. So p+np = - b/a, and np^2 = c/aOr (n+1) p = - b/a or p = - b/a (n+1) So n[-b/a (n+1) ]2 = c/aor nb2/a (n+1) 2 = cor nb2 = ac (n+1) 2

Which will give can^2 + (2ac-b^2) n + ac = 0, which is the required condition.