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Let alpha, beta, gamma be the roots of (x-a)(x-b)(x-c)=d, d not equal to 0, then the roots of the equation (x- alpha)(x- beta)(x- gamma)+d=0 are:(A) a+1,b+1,c+1(B) a,b,c(C) a-1,b-1,c-1(D) a/b,b/c,c/a

Agnik Das , 7 Years ago
Grade 11
anser 2 Answers
Arun

Last Activity: 7 Years ago

Dear student
 
(x -a) (x-b) (x -c) - d = (x-alpha) (x - beta) (x- gamma)
 
Now
 
(x -alpha) (x- beta) (x- gamma) + d = (x-a) (x-b) (x-c)
 
Hence the given equation has roota roota as
a, b, c
Ratnesh Asati

Last Activity: 4 Years ago

Dear friends,
In this question you will apply your brain rather than pen.
It is observed that I st equation is very similar to II nd 
Only d and -d are different but we know it is a clue to solve the question 
See here
 
                          \alpha + \beta +\gamma = a+b+c
       
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