If $a,b,c$ are three distinct positive real numbers then the number of real roots of $ax^2+2b|x|-c=0$ are

Question

The equation can have two forms $$ax^2+2bx-c+0$$ Or $$ax^2-2bx-c=0$$ both discriminants of the equation are $${4b^2+4ac}$$ which is a positive value so roots are real. Then there should be four roots of the given equation, but answer says there are only two. How is that possible?

Answer 1

Since $x^2 = |x|^2$ the given equation is same as:$$a|x|^2+2b|x|-c=0$$

This is a quadratic in $|x|$ and the quadratic formula gives $$|x|=(-b\pm \sqrt{b^2+ac})/a$$
However you must discard the negative root $(-b-\sqrt{b^2+ac})/a$ because $|x|$ is nonnegative.

$$|x| = (-b+\sqrt{b^2+ac})/a\Rightarrow x = \pm(-b+\sqrt{b^2+ac})/a$$

Answer 2

I got $$x_{1,2}=-\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}$$ or $$x_{1,2}=\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{b}{a}}$$ only $x_1$ (in both cases) are solutions.