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The product of all the roots of the eqn., $x^2-|x|-6=0$ is ?

$\begin{array}{1 1}(A) 6 \\ (B) -9 \\ (C) 36 \\(D)-6 \end{array}$

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  • If $x$ is real, then $ x^2=|x|^2$
Given eqn., is written as $|x|^2-|x|-6=0$
Solution of this equation is given by
$|x|=\large\frac{1\pm\sqrt {1+24}}{2}$
But $|x|$ cannot be negative.
$\therefore\:|x|=3\:\Rightarrow\:x=\pm 3$
Product of the roots $=-9$


answered Jul 25, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p

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