Question

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

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

Answer 1

The correct option is C a, b

Assume some convenient and appropriate values of a, b, c as
$a=3,b=4,c=6then(x-3)(x-4)-6=0\left[\right(x-a\left)\right(x-b)=c,c\ne 0]\Rightarrow x^2-7x+6=0\Rightarrow \alpha =6,\beta =1$
Again $(x-6)(x-1)+6(\because (x-\alpha \left)\right(x-\beta )+c=0)x^2-7x+6+6=0x^2-7x+12=0$
$\therefore$ the roots $k_1=3and{k}_2=4$
Which are same as a and b
Hence, option (c) is correct.

Alternatively:
$x^2-(a+b)x+(ab-c)=0\therefore \alpha +\beta =a+band\alpha \beta =ab-c$
Again if $k_1$ and $k_2$ be the roots of the other equation, then
$(x-\alpha )(x-\beta )+c=0i.e.,x^2-(\alpha +\beta )x+(\alpha \beta +c)=0\therefore k_1+k_2=\alpha +\beta =a+b\dots \left(i\right)and{k}_1.k_2=\alpha \beta +c-(ab-c+c=ab\dots (ii)$
Thus, from eq. (i) and (ii) it is clear that the roots are a and b
Hence correct choice is c.