Question

If $a\le 0$, then the roots of $x^2-2a|x-a|-3a^2=0$ is

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

Answer 1

Answer: (A)

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