If $a,b,c$ are three distinct positive real numbers, then the number of roots of:
$$ax^2+2b|x|-c=0$$
Please be very specific with the answer. The answer is $2$ but I think it should be $4$ because even if there are negative values of $x$ they will become $+ve$.
HINT: for $x\geq 0$ we have to solve $$ax^2+2bx-c=0$$ the Solutions are $$x_{1,2}=-\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}$$ and if $x<0$ you have to solve $$ax^2-2bx-c=0$$ $$x_{1,2}=\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}$$ in both case you will get the not the same Solutions