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$\alpha^2=5\alpha-3,\beta^2=5\beta-3\quad(\alpha\neq \beta)$,then the equation whose roots are $\large\frac{\alpha}{\beta}$ and $\large\frac{\beta}{\alpha}$ is :

$\begin{array}{1 1}(A)\;x^2-5x-3=0&(B)\;3x^2+12x+3=0\\(C)\;3x^2-19x+3=0&(D)\;None\end{array} $

1 Answer

$x^2-5x+3=0$
$\alpha+\beta=5$
$\alpha\beta=3$
$\large\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{19}{3}$
$\big(\large\frac{\alpha}{\beta}\big)\big(\large\frac{\beta}{\alpha}\big)$$=1$
Equation :$x^2-\large\frac{19}{3}$$x+1=0$
$\Rightarrow 3x^2-19x+3=0$
Hence (C) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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