Question

Let $\alpha, \beta, \gamma$ be the roots of $(x-a)(x-b)(x-c)=d, d \neq 0$, 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. $\frac{ a }{ b }, \frac{ b }{ c }, \frac{ c }{ a }$

Answer 1

Answer: (B)

Clearly $(x-a)(x-b)(x-c)-d$
$=(x-\alpha)(x-\beta)(x-\gamma)$
$\therefore$ if $\alpha, \beta, \gamma$ are the roots of given equation
then $(x-\alpha)(x-\beta)(x-\gamma)+d=0$
will have roots $a, b, c.$