Question

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

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

Answer 1

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

For $x\ge a$ ; $|x-a|=x-a$
$x^2-2a(x-a)-3a^2=x^2-2ax-a^2=0$
$x=\frac{2a\pm \sqrt{4a^2-4(+a{)}^2}}{2}$
$=\frac{2a\pm 2a\sqrt{2}}{2}$
$=(1+\sqrt{2})a$ for $x\ge a.$
For $x<a$ ; $(x-a)=-(x-a)$
$x^2+2a(x-a)-3a^2=0\Rightarrow x^2+2ax-5a^2=0$
$x=\frac{-2a\pm \sqrt{4a^2+20a^2}}{2}$
$=-a\pm \sqrt{6}a$
$=-(1+\sqrt{6})a.$
$\therefore$ The roots are $(1+\sqrt{2})a;-(1+\sqrt{6})a$.