Solving for $x$ (a quadratic equation)

Question

If $a\le0$, then roots of $x^2$ - $2a$|$x-a$| - $3a^2$ = $0$ are?

(A) $a$

(B) $($$\sqrt{6}$-$1$$)$$a$

(C) $($-$\sqrt{6}$-$1$$)$$a$

(D) None of these

I solved it by considering three cases.

Case-1 : When $x$ $>$ $a$ : In this case, $x$ = $a$ - $\sqrt{2}a$ is the root. (The other root $x$ = $a$ + $\sqrt{2}a$ gets rejected, since $x$ $>$ $a$)

Case-2 : When $x$ $<$ $a$ : In this case, $x$ = $($$\sqrt{6}$$-$$1$$)$$a$ is the root. (The other root $x$ = $($$-\sqrt{6}$$-$$1$$)$$a$ gets rejected since $x$ $<$ $a$)

Case-3 : When $x$= $a$ : In this case, I’m getting $a$ = $0$

The correct answer according to my book is (B).

(B) is indeed correct, but I don’t understand why $a$ = $0$ (Case-3) gets rejected? In that case I assumed that $x$ = $a$, and I got $a$ = $0$, which means $x$ = $a$ = $0$. And putting $x$ = $a$ = $0$ in the given equation reduces the LHS of the original equation to $0$. So $x$ = $a$ = $0$ should be a root too? This is the reason I am posting it. Shouldn’t $x$ = $a$ = $0$ be a root of the given equation?

Answer 1

Yes, $x = a$ is a root, but only for the case where $a = 0$, as you stated. However, note that for $a = 0$, the values of (A), (B) and (C) are also $0$. In particular, this case is already included in the case of $B$, which is possibly what the question had intended. Regardless, I consider it to be at least a misleading, if not also inaccurate, question.

Answer 2

Let $f_a(x):=x^2-2a|x-a|-3a^2.$

If $a=0$, then $f_a(x)=f_0(x)=x^2$ , hence $f_a(x)=0 \iff x=0.$

Conclusion: if $a=0$, then $f_a$ has the root $a$, and you are right.

Answer 3

1. If $a \geq 0$, then $a = 0$.
2. If $a=0$, then $x^2-2a|x-a|-3a^2=x^2$.
3. If $x^2-2a|x-a|-3a^2=0$, then $x^2=0$, because $x^2-2a|x-a|-3a^2=x^2$.
4. If $x^2=0$, then $x=0$.