IF C and D are the roots of `(x-a) (x-b) -k=0` then the roots of `(x-c) (x-d)+k=0` are

Let alpha,beta,gamma be the roots of (x-a) (x-b) (x-c) = d, d != 0 , then the roots of the equation (x-alpha)(x-beta)(x-gamma) + d =0 are :

Let alpha,beta,gamma be the roots of (x-a) (x-b) (x-c) = d, d != 0 , then the roots of the equation (x-alpha)(x-beta)(x-gamma) + d =0 are :

Let alpha,beta,gamma be the roots of (x-a) (x-b) (x-c) = d, d != 0 , then the roots of the equation (x-alpha)(x-beta)(x-gamma) + d =0 are :

Let alpha,beta,gamma be the roots of (x-a) (x-b) (x-c) = d, d != 0 , then the roots of the equation (x-alpha)(x-beta)(x-gamma) + d =0 are :

If c, d are the roots of the equation (x-a) (x-b) - k=0 , then prove that a, b are the roots of the equation (x-c) (x-d) + k= 0

If c and d are roots of the equation (x-a) (x-b) - k = 0, then a, b are roots of the equation

Let alpha,beta be the roots of the equation (x-a)(x-b)=c ,c!=0 Then the roots of the equation (x-alpha)(x-beta)+c=0 are a ,c (b) b ,c a ,b (d) a+c ,b+c

Let alpha,beta be the roots of the equation (x-a)(x-b)=c ,c!=0 Then the roots of the equation (x-alpha)(x-beta)+c=0 are a ,c (b) b ,c a ,b (d) a+c ,b+c