Question

Total number of integral values of a such that $x^2+ax+a+1=0$ has integral roots is equal to:

• A. one
• B. two
• C. three
• D. four

Answer 1

The correct option is A one

For a equation $a{x}^2+bx+c=0$ to have integer roots conditions are

$a-1,b\in I,c\in I$ and $b^2-4ac$ ia a perfect square

$x^2+ax+a+1=0$

$a\in I\therefore a+1\in I$

$a^2-4(a+1)$ is a perfect square

$a^2-4a-4=p^2⟹(a-2{)}^2-8-p^2=0⟹(a-2{)}^2=p^2+8$

$\therefore P^2=1$

$(a-2{)}^2=9⟹(a-2{)}^2=3^2$

$⟹a-2=3$ or $a-2=-3$

$a=5$ or $a=-1$

there are two values of a.

The number of integral values is one.