Question

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

• A. $(-1+\sqrt{6})a,a(1-\sqrt{2})$
• B. $(\sqrt{6}-1),(\sqrt{2}-1)$
• C. $a$
• D. none of these

Answer 1

The correct option is A $(-1+\sqrt{6})a,a(1-\sqrt{2})$

Case 1 : $x\ge a$ so $|x-a|=x-a$

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

$x^2-2ax+2a^2-3a^2=0$

$x^2-2ax=a^2$

$x^2-2ax+a^2=2a^2$

Taking square root,

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

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

But as a < 0, $a+a\sqrt{2}<a$

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

Case 2 : $x<a$ so $|x-a|=-(x-a)$

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

$x^2+2ax-2a^2-3a^2=0$

$x^2-2ax-5a^2=0$

By solving the equation, we get

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

but as $a<0,a(-1-\sqrt{6})>a.$

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

Therefore roots of equation are : $a(-1+\sqrt{6})$ and $a(1-\sqrt{2})$