Question

Integral value of $a$ such that quadratic equation $x^2+ax+a+1=0$ has integral roots is equal to

• A. $-1$
• B. $2$
• C. $1$
• D. $5$

Answer 1

Answer: (A), (D)

The correct options are
A $-1$
D $5$

Roots of the equation will be $\frac{-a\pm \sqrt{a^2-4a-4}}{2}$,

For integral roots condition ; $a^2-4a-4=k^2$ ,where k be any integer

$a^2-4a-(4+k^2)=0$

$a=2\pm \sqrt{8+k^2}$

Again the condition that $a$ is an integral value becomes $8+k^2=p^2$,where $p$ is an integral value.

We know that $x^2$ is an increasing function so the pair $(k,p)$ satisfying has a unique solution that is $(1,3)$.

$a=2\pm 3=-1,5$