Ia A, G and H are the arithmetic mean, geometric mean and harmonic mean between unequal positive integers. Then the equation ${\mathrm{Ax}}^2-\left|\mathrm{G}\right|\mathrm{x}-\mathrm{H}=0$ has

Given equation is ${\mathrm{Ax}}^2-\left|\mathrm{G}\right|\mathrm{x}-\mathrm{H}=0 ....\left(\mathrm{i}\right)$
$\begin{array}{l}\therefore \text{ Discriminant }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=(-|\mathrm{G}|{)}^2-4\mathrm{A}(-\mathrm{H})\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} ={\mathrm{G}}^2+4\mathrm{AH}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} ={\mathrm{G}}^2+4{\mathrm{G}}^2\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\because {\mathrm{G}}^2=\mathrm{AH}\right]\end{array}$
$\therefore \mathrm{Roots} \mathrm{of} \mathrm{Eq}.\; (\mathrm{i}) \mathrm{are} \mathrm{real} \mathrm{and} \mathrm{distinct}.$
$\because \mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}>0,\mathrm{G}=\sqrt{\mathrm{ab}}>0,\mathrm{H}=\frac{2\mathrm{ab}}{\mathrm{a}+\mathrm{b}}>0$
[$\because$a and b are two unequal positive integers]
Let $\mathrm{\alpha }\text{ and }\mathrm{\beta }$ be the roots of Eq. (i). Then, $\mathrm{\alpha }+\mathrm{\beta }=\frac{\left|\mathrm{G}\right|}{\mathrm{A}}>0$
and   $\mathrm{\alpha \beta }=-\frac{\mathrm{H}}{\mathrm{A}}<0$
and   $\mathrm{\alpha }-\mathrm{\beta }=\frac{\sqrt{\mathrm{D}}}{\mathrm{A}}=\frac{\mathrm{G}\sqrt{5}}{\mathrm{A}}>0$
$$\begin{gathered} \therefore \mathrm{\alpha }=\frac{\left|\mathrm{G}\right|+\mathrm{G}\sqrt{5}}{2\mathrm{A}}>0 \\ \mathrm{and} \mathrm{\beta }=\frac{\left|\mathrm{G}\right|-\mathrm{G}\sqrt{5}}{2\mathrm{A}}<0 \end{gathered}$$
Exactly one positive root and atleast one root which is negative fraction.