Question

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

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

Answer 1

Answer: (C)

Given $(x-a)(x-b)=c$
$\therefore \alpha +\beta =a+b$ and $\alpha \beta =ab-c$
Now. given equation $(x-\alpha )(x-\beta )+c=0$
$\Rightarrow x^2-(\alpha +\beta )x+\alpha \beta +c=0$
If its roots be $p$ and $q$, then
$p+q=(\alpha +\beta )=a+b$
$pq=\alpha \beta +c=ab-c+c=ab$
So, it can be given by $x^2-(a+b)x+ab=0$
So, its roots will be $a$ and $b$