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Let $A, G$ and $H$ be the arithmetic mean, geometric mean and harmonic mean, respetively of two distinct positive real numbers. If $\alpha$ is the smallest of the two roots of the equation $\mathrm{A}(\mathrm{G}-\mathrm{H}) \mathrm{x}^{2}+\mathrm{G}(\mathrm{H}-\mathrm{A}) \mathrm{x}+\mathrm{H}(\mathrm{A}-\mathrm{G})=0$, then
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Verified Answer
The correct answer is:
$0 < \alpha < 1$
$\mathrm{x}=1$ is a root as sum of coefficient $=0$
Now $\alpha \beta=\frac{H(A-G)}{A(G+H)}$
Put $\beta=1$
$\begin{aligned}
\alpha &=\frac{H A-H G}{A G-A H}=\frac{G^{2}-H G}{A G-A H} \\
&=\frac{G(G-H)}{A(G-H)}=\frac{G}{A} < 1 \quad[\text { as A.M. }>\text { G.M. }]
\end{aligned}$
Now $\alpha \beta=\frac{H(A-G)}{A(G+H)}$
Put $\beta=1$
$\begin{aligned}
\alpha &=\frac{H A-H G}{A G-A H}=\frac{G^{2}-H G}{A G-A H} \\
&=\frac{G(G-H)}{A(G-H)}=\frac{G}{A} < 1 \quad[\text { as A.M. }>\text { G.M. }]
\end{aligned}$
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