Question

If $(x+\alpha )$ is a common factor of $(x^2+ax+b)$ and $(x^2+cx+d)$ and $a:b:c:d=2:3:4:7,$ then the value of $\alpha$ is

Answer 1

Let $x^2+ax+b=0...\left(1\right)$
and $x^2+cx+d=0...\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right),$ we get
$(a-c)x+(b-d)=0$
$x=-\alpha$ is a root of both eqn$\left(1\right)$ and eqn$\left(2\right)$.
Therefore, putting $x=-\alpha$ in above equation, we get,
$\alpha =\frac{d-b}{c-a}$

Now, $a:b:c:d=2:3:4:7$
$\Rightarrow a=2k,b=3k,c=4k,d=7k$
$\therefore \alpha =\frac{7k-3k}{4k-2k}=2$