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The roots of $ax^{2}+bx+c = 0$ are real and positive. $a, b$ and $c$ are real. Then $ax^{2}+b\mid x \mid + c =0$ has 

  1. no roots
  2. $2$ real roots
  3. $3$ real roots
  4. $4$ real roots
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None of the answer is correct for all possible quadratic equations. Here is a counter example.

For $x^{2} - 2x + 1 = 0$, the only root is $1$ and it satisfies the criteria of being real & positive. Also the other criteria of  $a, b, c$ being real numbers is satisfied as well.

But even then the two equations $x^{2} - 2x + 1 = 0$, $x \geq 0$ and $x^{2} + 2x + 1 = 0$, $x < 0$ only have a total of 2 roots which are $1$ and $-1$ and not 4 roots.

All quadratic equations where the determinant is zero and $b$ is positive will only give 2 roots for such transformation.
Answer:

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