<s>[INST] How do i approach ahead in this question 
Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is

So what I did was,
I subtracted the two equations and got
$x=(a+b+2)/(a+b)$
I tried putting it in equation,it didn’t work,then  i tried adding the equations and then put the value,still didn’t work.
I cant seem to figure out how to approach this problem.Can anybody help out?
 [/INST] Let $q_a(x)=(a-1)x^2-(a^2+2)x+a^2+2a$. Any root $r$ of $q_a(x)$ and of $q_b(x)$ is also a root of$$(b-1)q_a(x)-(a-1)q_b(x)=\left(-a^2b+a^2+a b^2+2 a-b^2-2b\right)(x-1).$$Therefore, $r=1$ or$$-a^2b+a^2+a b^2+2 a-b^2-2b=0.\tag1$$But you can't have $r=1$, because\begin{align}1\text{ is a root of }q_a(x)&\iff q_a(1)=0\\&\iff3a-3=0\\&\iff a=1.\end{align}So, if $1$ was a root of both $q_a(x)$ and $q_b(x)$, you would have $a=b=1$, but you are assuming that $a\ne b$.
If, on the other hand, you have $(1)$, then $a=b$ or $a=\frac{b+2}{b-1}=1+\frac3{b-1}$. Since $a,b\in\Bbb N$, this can only occur in two case: when $b=2$ (in which case $a=4$) and when $b=4$ (in which case $a=2$). In both cases, you have $\frac{ab}5=\frac85$.
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