If one root of the quadratic equation px2+qx+r=0(p≠0) is aa+a−b, where p,q,r∈Q, and, a, b are positive integers which are not perfect square of any integer, then the other root is

One root aa+a−bOn rationalizing, we getaa+a−b×a−a−ba−a−b                    =a⋅(a−a−b)b=a−a(a−b)b∴  Other root =a+a(a−b)bRationalizing, we geta+a(a−b)b×a−a(a−b)a−a(a−b)            =a2−a2−abb(a−a(a−b))=aa−a−b