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 $x^2 +ax+b=0~~~...(1)$
and $x^2 +cx+d=0~~~...(2)$

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

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