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

Answer 1

The correct option is C $a,b$

$\because (x-a)(x-b)-c=(x-\alpha )(x-\beta )\therefore (x-\alpha )(x-\beta )+c=(x-a)(x-b)-c+c=(x-a)(x-b)$
Then the roots of $(x-\alpha )(x-\beta )+c=0$ are
$x=a,b$