माना समीकरण `(x-a)(x-b)=c,c ne 0` के मूल `alpha, beta` है, तो समीकरण `(x-alpha)(x-beta)+c=0` के मूल होंगे

If alpha, beta are the roots of (x-a) (x-b)+c=0, c ne 0 , then roots of (alpha beta - c) x^(2) + (alpha + beta) x + 1 = 0 are

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 :

If alpha and beta be two roots of the equation (x-a)(x-b)=c then show that a and b are the roots of (x-alpha)(x-beta) +c=0

If (x + 2)(x + 3b) = c has roots alpha,beta , then the roots of (x + alpha)(x+beta) + c = 0 are

If (x + 2)(x + 3b) = c has roots alpha,beta , then the roots of (x + alpha)(x+beta) + c = 0 are

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

If alpha and beta be the roots of the equation (x-a)(x-b)=c ( cne0 ),prove that a and b are the roots of the equation (x-alpha)(x-beta)+c=0 .

Let alpha , beta be the roots of the equation (x-a) (x-b) = c , cne 0 , then find the roots of the equation (x - alpha) (x -beta) + c = 0