If Alpha and beta are the roots of the equation ax^2 +bx +c =0. Then the roots of the equation a(x-2)^2 - b(x-2)(x-3) + c(x-3)^2= 0 is?

((Here Alpha=A, and Beta=B)) a(x-2)^2 - b(x-2)(x-3) + c(x-3)^2= 0Observe that this can be simplified as => (a-b+c)x^2 - (4a-5b+6c)x + (4a-6b+9c) =0Now divide all terms by `a`. So now you get terms like (-b/a) and (c/a) in the equation with their respective coefficients. Now we know that as ax^2+bx+c=0,1) (-b/a) = A+B 2) (c/a) =ABSo replace such terms in the equation using these... Now a, b, c have been completely eliminated. Now you get:=> (1+A+B+AB)x^2 - (4+5A+5B+6AB)x + (4+6A+6B+9AB)=0by the method of factorization and observation this becomes:(pq)x^2 + (rq+sp)x + (rs) =0Where,p=(1+A),q=(1+B),r=(2+3A),s=(2+3B).Using quadratic equation this can be easily solved to get the roots... Which are:1) (r/p) = (2+3A)/(1+A)2) (s/q) = (2+3B)/(1+B).

((Here Alpha=A, and Beta=B)).................................... a(x-2)^2 - b(x-2)(x-3) + c(x-3)^2= 0........................ Observe that this can be simplified as......... => (a-b+c)x^2 - (4a-5b+6c)x + (4a-6b+9c) =0...................Now divide all terms by `a`. .................So now you get terms like (-b/a) and (c/a) in the equation with their respective coefficients. .............Now we know that as ax^2+bx+c=0,.............(1) (-b/a) = A+B,,,,, (2) (c/a) =AB........................So replace such terms in the equation using these............. Now a, b, c have been completely eliminated. Now you get............=> (1+A+B+AB)x^2 - (4+5A+5B+6AB)x + (4+6A+6B+9AB)=0............................by the method of factorization and observation this becomes:...............(pq)x^2 + (rq+sp)x + (rs) =0...................Where,p=(1+A),q=(1+B),r=(2+3A),s=(2+3B)..................Using quadratic equation this can be easily solved to get the roots... ....Which are:........(1) (r/p) = (2+3A)/(1+A)..........(2) (s/q) = (2+3B)/(1+B).

Let as assume the roots to be x and y (alpha and beta) a(x-2)^2 - b(x-2)(x-3) + c(x-3)^2= 0 On simplifying, (a-b+c)x^2 - (4a-5b+6c)x + (4a-6b+9c) =0Now let`s divide each terms by `a`.So now we are getting terms (-b/a) and (c/a) in the equation with their coefficients. we know that,ax^2+bx+c=0_________(1) (-b/a) = A+B__________ (2) |||||(c/a) =AB So replace by replacing these terms in the equations, a, b, c are completely eliminated. Now we get::::::(1+A+B+AB)x^2 - (4+5A+5B+6AB)x + (4+6A+6B+9AB)=0By factorising__(pq)x^2 + (rq+sp)x + (rs) =0 {Where,p=(1+A),q=(1+B),r=(2+3A),s=(2+3B)}Using quadratic equation this can be easily solved to get the roots which are,__________(1) (r/p) = (2+3A)/(1+A)__________(2) (s/q) = (2+3B)/(1+B)Hope this is useful.....☺️