<s>[INST] How to find how many real roots of an equation? 
Roots of $ax^2 + bx + c = 0$ are real and positive. $a$, $b$ and $c$ are real.
Then $ax^2 + b|x| + c = 0$ has how many real roots?

My try:
I studied one method where we see how many signs are changing in equation. Then we are able to find real roots. But I forget the method name. So find difficult to solve.
Other methods are also appreciated. Thank you.
 [/INST] It's a weird question but it boils down to:  If $k > 0$ then $|k| = k$ and $|-k| = k$ and $(-k)^2 = k^2$ so if $k > 0$ is a solution to $ax^2 + bx + c$ then both $k$ and $-k$ are solutions to $ax^2 + b|x| + c$.
[Another way of putting it is $ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ so $k$ is a solution if and only if $-k$ is a solution.  If $x > 0$ then $a^2 + bx + c = ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ then $k$ is a positive solution to $a^2 + bx + c$ if and only if $k, -k$ are a positive and negative solution to $ax^2 + b|x| + c$..]
So if $a^2 + bx + c$ has two positive real solutions; $k,j$ then $a^2 + b|x| + c$ will have the same two positive real solutions, $k,j$ as well as their negative counter parts $-k,-j$ as solutions.
... Note: $ax^2 + b|x| + c$ is NOT a polynomial.  Hence the rule that there are at most two solutions does not apply.  (Actually, $ax^2 + b|x| + c$ is a system of two polynomials: $ax^2 + bx + c$ or $ax^2 - bx + c$ and as each may have up to two solutions, the system will have up to four solutions [to one or the other].  Furthermore, if $k,j$ are the solutions to $ax^2 + bx + c$ then $-k,-j$ are the solutions to $ax^2 -bx +c$.)
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