Question

The total number of integral value(s) of $a$ such that $x^2+ax+a+1=0$ has integral roots, is

• A. 2.0
• B. 2.00
• C. 2
• D. 02

Answer 1

Answer: (A), (B), (C), (D)

Let $\alpha$ and $\beta$ be the two integral roots of $x^2+ax+a+1=0.$
Then, $\alpha +\beta =-a\cdots \left(1\right)$
$\alpha \beta =a+1\cdots \left(2\right)$
Adding equations $\left(1\right)$ and $\left(2\right),$ we get
$\alpha +\beta +\alpha \beta =1$
$\Rightarrow (\alpha +1)(\beta +1)=2$
Since $\alpha$ and $\beta$ are integral roots
$\therefore \alpha +1=1,\beta +1=2$
$\Rightarrow \alpha =0,\beta =1$
or
$\alpha +1=-1,\beta +1=-2$
$\Rightarrow \alpha =-2,\beta =-3$
From equation $\left(1\right),$ we have
$a=-1,5$
Hence, total number of integral values of $a$ is $2.$