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If the roots of the equation $x^2-4x-log_3 a=0$ are real, then the least value of $a$ is ?

$\begin{array}{1 1}(A) \;81\\(B)\;64\\(C)\;\frac{1}{81}\\(D)\;\frac{1}{64}\end{array}$

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Since the roots of $x^2-4x-log_3 a=0$ are real,  Discriminent (D)$\geq 0$
$\Rightarrow\:16+4log_3 a\geq 0$
$\Rightarrow\:log_3 a\geq -4$
$\Rightarrow\:a\geq 3^{-4}$
$\therefore\:$ The least value of $a$ is $\large\frac{1}{81}$

 

answered Jul 14, 2013 by rvidyagovindarajan_1
edited Dec 19, 2013 by balaji.thirumalai
 

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