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 is where two curves of functions $y_1=3x^2+ax+1$ and $y_2=2x^2+bx+1$ intersect; so we may write:
$3x^2+ax+1=2x^2+bx+1$  ⇒  $x^2 +(a-b)x=0$   ⇒  $ x[x+(a-b)]=0$
$x=0$ is not acceptable
$x= b-a$ can be a solution; plugging this in one of equations we get:
$3(b-a)^2 +a(b-a) +1=0$
Reducing this relation we get:
$5ab -2a^2-3b^2=1$
