Question

Find ${\left(\frac{1}{10}\right)}^{th}$ of the sum of all integral values of $a$ for which the quadratic expression $(x-a)(x-10)+1$ can be factored as a product $(x+\alpha )(x+\beta )$ of two factors $\alpha ,\beta \in I$.

Answer 1

We have
$(x-a)(x-10)+1=(x+\alpha )(x+\beta )$
Putting $x=-\alpha$ in both sides, we obtain
$(-\alpha -a)(-\alpha -10)+1=0$

$(\alpha +a)(\alpha +10)=-1$

$\alpha +a$and $\alpha +10$are integer $(\because a,\alpha ϵI)$

$\therefore \alpha +a=-1$ and $\alpha +10=1$ or $\alpha +a=1$ and$\alpha +10=-1$
(i) If $\alpha +10=1$
$\therefore \alpha =-9$ then $a=8;$similarly, $\beta =-9$
Here, $(x-8)(x-10)+1={(x-9)}^2$
(ii) If $\alpha +10=-1$
$\therefore \alpha =-11$ then $a=12,$
Similarly, $\beta =12$

Here $(x-12)(x-10)+1={(x-11)}^2$.

$\therefore \frac{8+12}{10}=2$