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I need help proving that the roots of the equation: $(m-2)x^2-(3m-2)x+2m=0$ are real. Could you also give me a step by step runthrough of how to do this equation? (I have a few others like this one to do)

E.H.E
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Eidolon
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    From the quadratic form $Ax^2+Bx+C$, calculate the discriminant $$B^2-4AC$$ and make sure $B^2-4AC > 0$. Here, $$A=m-2, \qquad B=-(3m-2), \qquad C=2m$$ – Cookie Nov 17 '15 at 20:23
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    @Cookie: The condition for real roots is $\Delta\color{red}{\ge}0$. – Bernard Nov 17 '15 at 20:29

2 Answers2

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For $m=2$ we have a linear equation $-4x+4=0$, which has $x=1$ as the unique solution (assuming that $2\neq 0$ for the field). For $m\neq 2$ we have a quadratic equation, which has two roots, namely $$ x=\frac{2m}{m-2},\; x=1, $$ by the quadratic root formula. The first root may not be real in general, e.g., for $m=i\in \mathbb{C}$. If we assume $m\in \mathbb{R}$, both roots are real. If we assume that $m\in \mathbb{Z}$, both roots are rational.

Dietrich Burde
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The roots of the quadratic equation $ax^2+bx+c=0$ (with $a\ne0$) are real if and only if $b^2-4ac\ge0$; in your case we have \begin{align} b^2-4ac &=(3m-2)^2-8m(m-2)\\ &=9m^2-12m+4-8m^2+16m=m^2+4m+4\\ &=(m+2)^2 \end{align} This is never negative, so you're done, except for the case $m=2$, when the equation is not quadratic. But a linear equation with real coefficients has a real solution (if it is actually linear, of course).

Once you know that the discriminant is $(m+2)^2$, you can apply the quadratic formula (but it's not necessary to conclude the solutions are real): \begin{align} \frac{-b+\sqrt{b^2-4ac}}{2a}&=\frac{3m-2+m+2}{2(m-2)}=\frac{2m}{m-2} \\ \frac{-b-\sqrt{b^2-4ac}}{2a}&=\frac{3m-2-m-2}{2(m-2)}=1 \end{align} Note that $x=1$ is the only solution for the case $m=2$.

egreg
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