Question

The number of integral values of $a$ for which the quadratic equation $(x+a)(x+1991)+1=0$ has integral roots are

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

$(x+a)(x+1991)+1=0$

$\Rightarrow x^2+(a+1991)x+(1991a+1)=0$

For this equation,
$\Delta =(a+1991{)}^2-4(1991a+1)$

$\Delta =(a-1991{)}^2-4$

For integral roots,
$\Delta =(a-1991{)}^2-4=c^2$ , where $c$ is an integer

Let $k=a-1991$

$\Delta =k^2-4=c^2$

Now,
$k^2-c^2=4$

$(k-c)(k+c)=4$

Now, $k-c$ and $k+c$ will have the same parity.

Hence,
$k+c=2,k-c=2$ or $k+c=-2,k-c=-2$

$k=2$ or $k=-2$

Hence,
$a=1989$ or $a=1993$.

Hence, there are two values of $a$.