Explanation

(b)
Case I When $n \geq a$
$\therefore\ \ \ x^{2} - 2a(x - a) - 3a^{2} = 0$
$\Rightarrow \ \ x^{2} - 2ax - a^{2} = 0\ \ \Rightarrow \ \ x = a \pm \sqrt{2}a$
Now, for $x \geq a,\ a < 0$
$\Rightarrow \ \ \ \ \ x = a\left( 1 - \sqrt{2} \right)\ \ \ \lbrack\because x = a\left( 1 + \sqrt{2} \right) < a\rbrack$ ...(i)
Case II When $x < a$
$\therefore\ \ x^{2} + 2a(x - a) - 3a^{2} = 0$
$\Rightarrow \ \ x^{2} + 2ax - 5a^{2} = 0$
$\Rightarrow \ \ \ x = - a \pm \sqrt{6}a$
Now, for $x < a,\ \ \ a < 0$
$\Rightarrow \ \ \ \ x = a(\sqrt{6} - 1)$ ...(ii)
$\left\lbrack \because\ \ x = - a\left( 1 + \sqrt{6} \right) > a \right\rbrack$
From Eqs. (i) and (ii),
$x = \left\{ a\left( 1 - \sqrt{2} \right),\ a\left( \sqrt{6} - 1 \right) \right\}$