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Given the equation: $$(3m-5)x^2-3m^2x+5m^2=0$$ Prove the roots are rational given $m$ is rational.

I have tried finding the discriminate of the quadratic however this has not been fruitful and ends up being very ugly. I have tried setting the discriminate to $a^2$ however I don't know if this is the right thing to do.

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  • $\begingroup$ (3m-5)x^2... or (3m^2-5)x^2...? $\endgroup$ – mrs Sep 21 '17 at 10:27
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Why, just calculate the discriminant! $$\Delta=(-3m^2)^2-4(3m-5)(5m^2)=9m^4-60m^3+100m^2=(3m^2-10m)^2$$ and so $\sqrt\Delta=|3m^2-10m|$. The remainder of the quadratic formula consists of rational manipulations, so the roots are rational if $m$ is rational.

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  • $\begingroup$ @JeanMarie Noted. $\endgroup$ – Parcly Taxel Sep 21 '17 at 10:52
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Yes, you can calculate of course the discriminant and to end this problem,

but I think it's better to use the Viet's theorem.

If $a\neq0$ and $x_1$, $x_2$ are roots of the equation $$ax^2+bx+c=0$$ then $$x_1+x_2=-\frac{b}{a}$$ and $$x_1x_2=\frac{c}{a}.$$

Indeed, easy to see that $m$ is a rational root because $(3m-5)m^2- 3m^2\cdot m+5m^2=0.$ Thus, for $m\neq\frac{5}{3}$ we get more rational root: $\frac{5m}{3m-5}$ because $x_1x_2=\frac{5m^2}{3m-5}.$

If $m=\frac{5}{3}$ then our equation has one rational root and we are done!

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You could always spot that $x=m$ is a solution.

To get this you can use the rational root theorem, and note that the coefficient of $x^2$ has no factors - if the roots are rational for every $m$, therefore, there should be a factorisation $\left(\left(3m-5\right)x+a\right)\left(x+b\right)$ and $b$ must be a factor of $5m^2$ so there is not far to look.

Otherwise just use the quadratic formula to find the roots.

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