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?

If you like using Vieta's formulas, here is the possible solution based on these formulas:
Assuming $x$ as a coefficient, you can observe that, $a$ and $b$ are root of the quadratic respect to the variable $u:$
$$u^2(1-x)+u(x^2+2)-(x^2+2x)=0, ~ x≠1$$
Then, using the Vieta's formulas, we have
$$\begin{align}&ab=\frac{x^2+2x}{x-1}\\
\implies &x^2+x(2-ab)+ab=0 \\
\implies &x_1x_2=ab\end{align}$$
Then, we also have a quadratic respect to the variable $x:$
$$(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0, ~ a≠1$$
Since $a≠0$, applying Vieta's formulas again,  we get
$$\begin{align}&\frac{a^2+2a}{a-1}=ab\\
\implies &b=\frac{a+2}{a-1}\\
\implies &b=1+\frac{3}{a-1}\\
\implies &a\in\left\{2,4\right\}\\
\implies &ab=8 \\
\implies &\frac{ab}{5}=\frac 85. \end{align}$$
