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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 ?

Victor
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2 Answers2

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

Bill Dubuque
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  • That was really stupid of me, not noticing such a trivial trick. Thanks a lot – Victor Aug 25 '14 at 21:17
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    @victor We've all overlooked simple things. When that occurs, I recommend performing some meta-reflection to attempt to understand the source of the oversight. E.g. above, perhaps it was not obvious that one could eliminate the expression $,1/m,$ just like one can eliminate a variable. If you constantly perform such meta-level reflection, it can go a long way towards strengthening your problem-solving skills. – Bill Dubuque Aug 25 '14 at 22:16
  • That's perfectly true Mr. Dubuque, thanks a lot for your extremely useful input ! – Victor Aug 25 '14 at 23:25
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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}

Lythia
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