Question

The number of integral values of $a$ for which the quadratic equation $(a+2)x^2+2(a+1)x+a=0$ will have integer roots are

• A. $3$
• B. $2$
• C. $5$
• D. $4$

Answer 1

Answer: (D)

$4$

Let us calculate the discriminant of the given quadratic equation.

$D=b^2-4ac$

$D={\left(2(a+1)\right)}^2-4a(a+2)$
$D=4a^2+8a+4-4a^2-8a$

$D=4$

So, the roots of the given equation will be,

$x=\frac{-b\pm \sqrt{D}}{2a}$
$x=\frac{-2(a+1)\pm 2}{2(a+2)}$

$x=-\frac{a}{a+2},-1$

As $x=-1$ is already an integral solution.

So, we will consider the other case.

$x=-\frac{a}{a+2}$

$=-\frac{a+2-2}{a+2}$

$=-1+\frac{2}{a+2}$

For $\frac{2}{a+2}$ to be an integer, the possible values of $a$ are $0,-1,-3$ and $-4.$
Hence, there are $4$ integral values of $a$.