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If the eqns., $x^2-6x+a=0$ and $x^2-5x+6=0$ have one root in common and the other roots are in the ratio 4:3, then the common root is

$\begin{array}{1 1} 1 \\ 2\\ 3 \\ 4 \end{array}$

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The given eqns., $x^2-6x+a=0.....(i),\:x^2-5x+6=0.....(ii)$
Let the roots of (i) be $\alpha\:\:and\:\:\beta\:\: and$
that of $(ii)$ be $\alpha\:\:and\:\:\gamma$
$\Rightarrow\:\alpha+\beta=6,\:\alpha\beta=a,\:\alpha+\gamma=5\:and\:\alpha\gamma=6$
Also given that $\beta:\gamma=4:3$
$\Rightarrow\:\large\frac{\alpha\beta}{\alpha\gamma}=\frac{a}{6}$
$\Rightarrow\:\large\frac{\beta}{\gamma}=\frac{a}{6}=\frac{4}{3}$
$\Rightarrow\:a=8$
$\Rightarrow\:\alpha\beta=a=8$
$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$
$\Rightarrow\:(\alpha-\beta)^2=36-32=4$
$\Rightarrow\:\alpha-\beta=\pm 2$
$\alpha+\beta=6$
$\Rightarrow\:\alpha=4\:or\:2$
But $\alpha=4$ does not satisfy the 2nd eqn.,
$\therefore\:\alpha=common\:root=2$
answered Aug 5, 2013 by rvidyagovindarajan_1
edited May 28, 2014 by rohanmaheshwari0831_1
 

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