$ax^{2} + bx + c$, for roots to be real & +ve $b^{2}$ − 4ac > 0
This will have 2 real positive roots.
$ax^{2} + b|x| + c$ = 0
This can be written as
i) $ax^{2} + bx + c$ = 0 (x>=0)
Discriminant = $b^{2}$ − 4ac > 0. --> Roots are real and positive
And,
ii) $ax^{2} - bx + c$ = 0 (x<0)
$(-b)^{2}$ − 4ac ⇒ $b^{2}$ − 4ac is also >0. --> Roots are real and positive
So, we will have 4 real roots for $ax^{2} + b|x| + c$ = 0.
Therefore, correct answer is option (D).