Finding value of Quadratic

Question

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?

Answer 1

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*}

Answer 2

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:

$5ab -2a^2-3b^2=1$