Question

If $a<0$, the positive root of the equation $x^2-2a$ $|x-a|-3a^2=0$ is

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

Answer 1

Answer: (C)

Since $a<0$, in case of positive root of the equation $x>a$
$\therefore$ The equation is $x^2-2a(x-a)-3a^2=0$
$\Rightarrow x^2-2ax-a^2=0$
Thus, the roots are $\frac{2a\pm \sqrt{4a^2+4a^2}}{2}$
$=\frac{2a\pm 2\sqrt{2}a}{2}$ $=a(1+\sqrt{2})$
or $a(1-\sqrt{2})$
$\therefore$ the only positive root possible is $a(1-\sqrt{2})$.