Question

For $a\le 0$, the roots of the equation $x^2-2a\left|x-a\right|-3a^2=0$ is

• A. $-a\left(\sqrt{6}+1\right)$
• B. $a\left(\sqrt{6}+1\right)$
• C. $a\left(1-\sqrt{2}\right),a\left(-1+\sqrt{6}\right)$
• D. $a\left(1-\sqrt{2}\right)$

Answer 1

The correct option is D $a\left(1-\sqrt{2}\right)$

$x^2-2a|x-a|-3a^2=0$

Case $I$

$x\ge a$

$⟹x^2-2a(x-a)-3a^2=0$

$⟹x^2-2ax+a^2=2a^2$

$⟹x-a=\pm \sqrt{2}a$

$⟹x=a\pm \sqrt{2}a$

But, $a<0$

$\therefore x=a(1-\sqrt{2})$

Case $II$

as, $x<0$, $a<0$

But we want only positive roots, we have nothing to do with this case.

Thus, $x=a(1-\sqrt{2})$