If $a, b, c$ are three distinct positive real numbers, then the number of real root(s) of $a x^{2} + 2 b | x | + c = 0$ is

0

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2

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Solution

The correct option is A $0$
The given equation can be rewritten as
$a | x |^{2} + 2 b | x | + c = 0$
We know that $a, b, c > 0$ and all are distinct.
Using quadratic formula
$| x | = \frac{- 2 b \pm \sqrt{4 b^{2} - 4 a c}}{2 a}$

Case $1 :$
$b^{2} - 4 a c < 0$
$\Rightarrow$ No real solution
Case $2 :$
$b^{2} - 4 a c > 0$
$4 b^{2} > 4 b^{2} - 4 a c \Rightarrow 2 b > \sqrt{4 b^{2} - 4 a c} \Rightarrow - 2 b + \sqrt{4 b^{2} - 4 a c} < 0 \Rightarrow \frac{- 2 b + \sqrt{4 b^{2} - 4 a c}}{2 a} < 0$
And $\frac{- 2 b - \sqrt{4 b^{2} - 4 a c}}{2 a} < 0$

Therefore $| x | = some negative value$ ,
which is not possible.
Hence, the given equation has no real solution.

Alternate Solution :
$a x^{2} + 2 b | x | + c = 0$
$x^{2} \geq 0, | x | \geq 0$
Given $a, b, c > 0$
$\Rightarrow L.H.S. > 0$ but $R.H.S. = 0$
Hence, the given equation has no real solution.

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