Question

If $\alpha ,\beta ,\gamma$ 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

The correct option is C $a,b,c$

$\because \alpha ,\beta ,\gamma$ are the roots of the given equation,
$(x-a)(x-b)(x-c)-d=(x-\alpha )(x-\beta )(x-\gamma )\Rightarrow (x-a)(x-b)(x-c)=(x-\alpha )(x-\beta )(x-\gamma )+d$

$\therefore$ Roots of the equation $(x-\alpha )(x-\beta )(x-\gamma )+d=0$ are $a,b,c$