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 $a,b$

$(x-a)(x-b)-c=0$
$x^2-(a+b)x+ab-c=0$
Hence
$\alpha .\beta =ab-c$
and
$\alpha +\beta =a+b$
Now
$(x-\alpha )(x-\beta )+c=0$
$x^2-(\alpha +\beta )x+\alpha .\beta +c=0$
$x^2-(a+b)x+ab-c+c=0$
$x^2-(a+b)x+ab=0$
$(x-a)(x-b)=0$
$x=a$ and $x=b$