Question

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