Finding value of Quadratic If the quadratic equations $3x^2+ax+1=0$ and $2x^2+bx+1=0$ have a common root, then the value of $5ab-2a^2-3b^2$ has to be find.
I tried by eliminating the terms but ended with $(2a-b)x=1$. Can you please suggest how to proceed further?

The common root, $r$, is also root of 
\begin{align*}
3x^2+ax+1-(2x^2+bx+1)&=0\\
x^2+(a-b)x&=0\\
x(x+a-b)&=0
\end{align*}
The root cannot be $zero$, so we get $r=b-a$ as the common root.
Also, the common root is a root of
\begin{align*}
2(3x^2+ax+1)-3(2x^2+bx+1)&=0\\
(2a-3b)x-1&=0\\
x&=\frac1{2a-3b}\implies r=\frac1{2a-3b}
\end{align*}
Now,
\begin{align*}
5ab-2a^2-3b^2&=(b-a)(2a-3b)\\
&=r\left(\frac1r\right)\\
&=1
\end{align*}
