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Q.
If the ratio of the roots of $\lambda x^{2}+\mu x+v=0$ is equal to the ratio of the roots of $x^{2}+x+1=0$, then $\lambda, \mu, v$ are in
Complex Numbers and Quadratic Equations
Solution:
Let $\alpha, \beta$ be the roots of $\lambda x^{2}+\mu x+v=0$
$\therefore \alpha+\beta=-\frac{\mu}{\lambda}, a \beta=\frac{v}{\lambda}$
$\therefore \frac{(\alpha+\beta)^{2}}{\alpha \beta}=\frac{\frac{\mu^{2}}{\lambda^{2}}}{\frac{v}{\lambda}}=\frac{\mu^{2}}{\lambda v} $
$\Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2=\frac{\mu^{2}}{\lambda v} \,\,\,\, (1)$
Let $\gamma, \delta$ be the roots of $x^{2}+x+1=0$
$\therefore \gamma+\delta=-1, \gamma \delta=1 $
$\therefore \frac{(\gamma+\delta)^{2}}{\gamma \delta}=1 $
$\Rightarrow \frac{\gamma}{\delta}+\frac{\delta}{\gamma}+2=1 \,\,\,\, (2)$
Since $\frac{\alpha}{\beta}=\frac{\gamma}{\delta}$
$\therefore $ from Eq. (1) and (2)
$\frac{\mu^{2}}{\lambda v}=1 $
$\Rightarrow \mu^{2}=\lambda v$
$\therefore \lambda, \mu, v$ are in G.P.