Question

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

• A. a$(1+\sqrt{2})$, a$(\sqrt{6}-1)$ if a < 0
• B. 0, 0 if a=0
• C. a$(1-\sqrt{2})$, a$(1-\sqrt{6})$ if a < 0
• D. a$(1-\sqrt{2})$, a$(\sqrt{6}+1)$ if a < 0

Answer 1

The correct option is C a$(1-\sqrt{2})$, a$(1-\sqrt{6})$ if a < 0

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

Case I: When $x-a\ge 0$
$x^2-2a(x-a)-3a^2=0$
$\Rightarrow x=\frac{2a\pm 2\sqrt{2}a}{2}$
$\Rightarrow x=a(1\pm \sqrt{2})$

Since, $a\le 0$
For real roots , $D\ge 0$
$\Rightarrow x=a(1-\sqrt{2})$
If $a=0$ ,$x=0$

So, in first case we have 2 roots $x=0$ if $a=0$ and $x=a(1-\sqrt{2})$ if $a<0$
Case II: When $|x-a|<0$
$x^2+2a(x-a)-3a^2=0$
$\Rightarrow x=\frac{2a\pm 2\sqrt{6}a}{2}$
$\Rightarrow x=a(1\pm \sqrt{6})$

Since, $a\le 0$

For real roots , $D\ge 0$
$\Rightarrow x=a(1-\sqrt{6})$
If $a=0$ ,$x=0$

So, in second case we have 2 roots $x=0$ if $a=0$ and $x=a(1-\sqrt{6})$ if $a<0$