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If one root of the equation $8x^2-6x-k-3=0$ is square of other,then value of $k$ is

$\begin{array}{1 1}(A)\;4,24&(B)\;-4,-24\\(C)\;4,-24&(D)\;-4,24\end{array} $

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Let the roots be $\alpha,\alpha^2$
$\alpha+\alpha^2=\large\frac{6}{8}$$\Rightarrow \alpha(1+\alpha)=\large\frac{3}{4}$$\Rightarrow \alpha=\large\frac{-3}{2},\frac{1}{2}$
$\alpha^3=-\large\frac{(k+3)}{8}$
$\Rightarrow -\large\frac{(k+3)}{8}=\frac{-27}{8},\frac{1}{8}$
$\Rightarrow -(k+3)=-27,1$
$\Rightarrow k+3=27,-1$
$\Rightarrow k=24,-4$
Hence (D) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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