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The numerical difference of the roots of $x^2-6x+6=0$ is ?

$\begin{array}{1 1}(A) 0 \\ (B) \sqrt {6} \\ (C) \sqrt {12} \\(D) \sqrt {18} \end{array}$

1 Answer

Given: $x^2-6x+6=0$
$\Rightarrow\:x=\large\frac{6\pm \sqrt{36-24}}{2}=\frac{6\pm\sqrt{12}}{2}$
$\Rightarrow\:3+\sqrt 3\:\:or\:\:3-\sqrt 3$
Difference of the roots $=2\sqrt 3=\sqrt {12}$
answered Jul 27, 2013 by rvidyagovindarajan_1
 

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