Question

For $a\le 0$, determine all real roots of the equation $x^2-2a|x-a|-3a^2=0$.

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

Answer 1

Answer: (A), (C)

$(\sqrt{6}-1)a$
$(1-\sqrt{2})a$

Here, $a\le 0$

Given,

Case I. When $x\ge a$

$\Rightarrow x^2-2a(x-a)-3a^2=0$

$\Rightarrow x^2-2ax-a^2=0$

$\Rightarrow x=a\pm \sqrt{2}a$

[ as, $a(1+\sqrt{2})<a$ and $a(1-\sqrt{2}>a)$]

$\therefore$ Neglecting $x=a(1+\sqrt{2})$ as $x\ge a$

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

Case (II)

When $x<a\Rightarrow x^2+2a(x-a)-3a^2=0$

$\Rightarrow x^2+2ax-5a^2=0\Rightarrow x=-a\pm \sqrt{6}a$

[as, $a(\sqrt{6}-1)<a$ and $a(1-\sqrt{6})>a$]

$\therefore$ neglecting $x=a(-1-\sqrt{6})\Rightarrow x=a(\sqrt{6}-1)$ ...(2)

From Eqs. (1) and (2), we get

$x=\{a(1-\sqrt{2}),a(\sqrt{6}-1)\}$