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Let $\alpha, \beta, \gamma(\alpha < \beta < \gamma)$ be roots of $a x^3+b x^2+c x+d=0$ and $u, v, w(u < v < w)$ be roots of $a k^3 x^3+b k^2 x^2+c k x+d=0$. If $\beta^2=\alpha \gamma$, then
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Verified Answer
The correct answer is:
$v^2=u w$
$\because \alpha, \beta$ and $\gamma$ are the roots of $a x^3+b x^2+c x+d=0$ and given $\beta^2=\alpha \gamma$ i.e. $\alpha, \beta$ and $\gamma$ are in G.P.
So, $\frac{\alpha}{K}, \frac{\beta}{K}$ and $\frac{\gamma}{K}$ are also in G.P.
Now, $u, v$ and $w$ are roots of
$\begin{aligned}
a k^3 x^3+b k^2 x^2+c k x+d & =0 \\
\text { i.e. } \quad a(k x)^3+b(k x)^2+c(k x)+d & =0 \quad \ldots
\end{aligned}$
So, roots of Eq. (i) are $\frac{\alpha}{K}, \frac{\beta}{K}$ and $\frac{\gamma}{K}$ i.e. $u=\frac{\alpha}{K}$, $v=\frac{\beta}{K}$ and $w=\frac{\gamma}{K}$
From above, $u, v$ and $w$ are also in G.P. Hence, $v^2=u w$.
So, $\frac{\alpha}{K}, \frac{\beta}{K}$ and $\frac{\gamma}{K}$ are also in G.P.
Now, $u, v$ and $w$ are roots of
$\begin{aligned}
a k^3 x^3+b k^2 x^2+c k x+d & =0 \\
\text { i.e. } \quad a(k x)^3+b(k x)^2+c(k x)+d & =0 \quad \ldots
\end{aligned}$
So, roots of Eq. (i) are $\frac{\alpha}{K}, \frac{\beta}{K}$ and $\frac{\gamma}{K}$ i.e. $u=\frac{\alpha}{K}$, $v=\frac{\beta}{K}$ and $w=\frac{\gamma}{K}$
From above, $u, v$ and $w$ are also in G.P. Hence, $v^2=u w$.
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