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If one root of $(a^2-5a+3)x^2+(3a-1)x+2=0$ is double the other,then the value of a is :

$\begin{array}{1 1}(A)\;\large\frac{2}{3}&(B)\;\large\frac{1}{3}\\(C)\;\large\frac{-1}{3}&(D)\;\large\frac{-2}{3}\end{array} $

1 Answer

Let the roots be $\alpha,2\alpha$
$(\alpha(2\alpha)=\large\frac{2}{a^2-5a+3}$$\Rightarrow \alpha^2=\large\frac{1}{a^2-5a+3}$
$\alpha+2\alpha=3\alpha=-\large\frac{-(3a-1)}{a^2-5a+3}$
$\Rightarrow 3\alpha=(1-3a)\alpha^2\Rightarrow \alpha=\large\frac{3}{1-3a}$
$2\alpha=\large\frac{6}{1-3a}$
$\alpha^2=\large\frac{9}{(1-3a)^2}=\frac{1}{a^2-5a+3}$
$\Rightarrow 9a^2-45a+27=9a^2-6a+1$
$\Rightarrow 26=39a\Rightarrow a=\large\frac{2}{3}$
Hence (A) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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