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Q.
If $a \le\,0$, then the roots of $x^{2}-2a |x-a|-3a^{2}=0$ is
Complex Numbers and Quadratic Equations
Solution:
When $x-a<\,0$ then $|x-a| = -(x-a)$
$\therefore $ equation $x^{2}+2a(x-a) -3a^{2}=0$
or $x^{2}+2ax-5a^{2}=0$
$\therefore \left(x+\left(1+\sqrt{6}\right)a\right) \left(x+\left(1-\sqrt{6}\right)a\right)=0$
$\therefore x=-\left(1+\sqrt{6}\right)a$ or $x=\left(-1+\sqrt{6}\right)a$
$\because x <\,a \le\,0$
therefore, only $x= \left(-1 +\sqrt{6}\right)$ a is possible
$\therefore $ Option $\left(a\right)$ is correct
when $x-a \ge\,0$ then $\left|x-a\right|=\left(x-a\right)$
$\therefore $ quation is $x^{2}-2a \left(x-a\right)-3a^{2}=0$
$\Rightarrow x^{2}-2ax -a^{2}=0 $
$\Rightarrow \left[x-\left(1+\sqrt{2}\right)a\right]\left[x+\left(-1+\sqrt{2}\right)a\right]=0$
$\Rightarrow x=\left(1+\sqrt{2}\right)a $
or $x=\left(1-\sqrt{2}\right)$ a also lies in the range but not given in the options