Question

If $a\le 0$ then the real roots of the equation $x^2-2a|x-a|-3a^2=0$ is/are

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

Answer 1

Answer: (B), (C)

$-a+a\sqrt{6}$

The given equation is
$x^2-2a|x-a|-3a^2=0$
Here, Two cases are possible

Case (1)$x-a>0$ then equation becomes
$x^2-2a(x-a)-3a^2=0$
$\Rightarrow x^2-2ax-a^2=0$
$\Rightarrow x=\frac{2a\pm \sqrt{4a^2+4a^2}}{2}$
$\Rightarrow x=a\pm a\sqrt{2}$
As $a\le 0$
So, $x=a-a\sqrt{2}$

Case (2)$x-a<0$, then equation becomes
$x^2+2a(x-a)-3a^2=0$
$\Rightarrow x^2+2ax-5a^2=0$
$\Rightarrow x=\frac{-2a\pm \sqrt{4a^2+20a^2}}{2}$
$\Rightarrow x=-a\pm a\sqrt{6}$
As $a\le 0$
So, $x=-a+a\sqrt{6}$

Thus the solution set is $\{a-a\sqrt{2},-a+a\sqrt{6}\}$