Problem involving a quadratic

Question

This is the equation : $mx^2-(4m-1)x+3m-2=0 $

We are asked to find a relationship between the roots that doesn't involve the 'm' parameter. Therefore, I thought the key to achieving this was making use of Vieta's formulas, but nothing useful came out no matter how much I played with the equations .

Can someone point me to the solution ?

Answer 1

$\begin{eqnarray}{\bf Hint}\qquad\qquad r+s &=& 4-1/m\\ r\,*\,s &=& 3-2/m\\ \hline \\ \Rightarrow\ 2(r+s)-r*s &=& \ \ldots\qquad \text{by eliminating}\,\ 1/m \end{eqnarray}$

Answer 2

Let $x_1$ and $x_2$ be the two roots of the given quadratic equation. Vieta's formulas give that \begin{align} x_1+x_2 &= \frac{4m-1}{m},\\ x_1x_2 &= \frac{3m-2}{m}. \end{align} Multiplying the first equation by $2m$ and the second equation by $m$ gives \begin{align} 2m(x_1+x_2) &= 8m-2,\\ mx_1x_2 &= 3m-2. \end{align} Subtracting the second equation from the first gives that \begin{equation} m(2(x_1+x_2)-x_1x_2) = 5m. \end{equation} Therefore the $m$ cancels, leaving \begin{equation} 2(x_1+x_2)-x_1x_2 = 5. \end{equation}