Question

Let $\alpha ,\beta .\gamma$ be the roots of the equation $(x-a)(x-b)(x-c)=d,d\ne 0$, then the roots of the equation $(x-\alpha )(x-\beta )(x-\gamma )+d=0$, are

• A. $a,b,d$
• B. $b,c,d$
• C. $a,b,c$
• D. $a+d,b+d,c+d$

Answer 1

Answer: (C)

$a,b,c$

Since $\alpha$,$\beta$ and $\gamma$ are the roots of the equation$(x-a)(x-b)(x-c)-d=0$.
So we have
$\alpha +\beta +\gamma =a+b+c$ ...(1)
$\alpha \beta +\beta \gamma +\gamma \alpha =ab+bc+ca$ ...(2)
$\alpha \beta \gamma =abc+d$ ...(3)
Similarly, for the equation $(x-\alpha )(x-\beta )(x-\gamma )+d=0$, we have
$\alpha +\beta +\gamma =a+b+c$ (from (1))
$\alpha \beta +\beta \gamma +\gamma \alpha =ab+bc+ca$ (from (2))
And $\alpha \beta \gamma -d=abc$ (from (3))
Therefore clearly $a$,$b$ and $c$ will be the roots of the equation $(x-\alpha )(x-\beta )(x-\gamma )+d=0$.