If a, b, c are three distinct positive real numbers, then the number of real roots of \[a{{x}^{2}}+2b\left| x \right|-c=0\] is
\[\begin{align}
  & \text{A}.\text{ 4} \\
 & \text{B}.\text{ 2} \\
 & \text{C}.\text{ }0 \\
 & \text{D}.\text{ None of above} \\
\end{align}\]

Hint: When the quadratic equation is \[a{{x}^{2}}+bx+c=0\] then the roots of it is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where the part \[D={{b}^{2}}-4ac\] is called discriminant D of the quadratic equation. If D>0 or D=0, we will have real roots. To solve this question, we will use the above stated formula to find the roots of the equation first. We will also open the modulus function and we know that it will have both positive and negative values. Then, we will examine which of the obtained roots are actually possible, then we would get the number of roots from it.

Complete step-by-step answer:
We have to find number of real roots of equation \[a{{x}^{2}}+2b\left| x \right|-c=0\]
We can clearly write \[{{x}^{2}}\text{ as }{{\left| x \right|}^{2}}\]
\[a{{\left| x \right|}^{2}}+2b\left| x \right|-c=0\]
Let us assume $y=\left| x \right|$ then equation becomes; \[a{{\left| x \right|}^{2}}+2b\left| x \right|-c=0\Rightarrow a{{y}^{2}}+2by-c=0\] the root of equation can be given by using the formula to find roots of quadratic equation which is as below;
Here we are given equation as
\[a{{y}^{2}}+2by-c=0\]
Using the formula stated above, we get:
\[\begin{align}
  & y=\dfrac{-2b\pm \sqrt{{{\left( 2b \right)}^{2}}+4\times ac}}{2a} \\
 & y=\dfrac{-2b\pm \sqrt{4{{b}^{2}}+4ac}}{2a} \\
\end{align}\]
Taking 2 common from both numerator and denominator, we get:
\[\begin{align}
  & y=\dfrac{\left( -b\pm \sqrt{{{b}^{2}}+ac} \right)}{a} \\
 & y=\dfrac{-b\pm \sqrt{{{b}^{2}}+ac}}{a} \\
\end{align}\]
So, there are two possibility
\[y=\dfrac{-b+\sqrt{{{b}^{2}}+ac}}{a}\text{ and }y=\dfrac{-b-\sqrt{{{b}^{2}}+ac}}{a}\]
We had $y=\left| x \right|$
\[\left| x \right|=\dfrac{-b+\sqrt{{{b}^{2}}+ac}}{a}\text{ and }\left| x \right|=\dfrac{-b-\sqrt{{{b}^{2}}+ac}}{a}\]
As the value of $\left| x \right|$ is always positive for any x, so the term \[\left| x \right|=\dfrac{-b-\sqrt{{{b}^{2}}+ac}}{a}\] is always positive.
LHS of above is positive but RHS of above has both terms negative, so it is negative, which is not possible.
Hence, \[\left| x \right|=\dfrac{-b-\sqrt{{{b}^{2}}+ac}}{a}\] is not possible.
Hence the left possibility is \[\left| x \right|=\dfrac{-b+\sqrt{{{b}^{2}}+ac}}{a}\]
When any number a is written as $\left| a \right|=t$ where t is any number, then we open a as $a=\pm t$. Applying this in above equation, we get:
\[x=\dfrac{\pm \left( -b+\sqrt{{{b}^{2}}+ac} \right)}{a}\]
Again, we have two possibility:
\[x=\dfrac{+\left( -b+\sqrt{{{b}^{2}}+ac} \right)}{a}\Rightarrow x=\dfrac{-\left( -b+\sqrt{{{b}^{2}}+ac} \right)}{a}\]
Therefore, there are two roots possible which are real. Hence, we have
\[x=\dfrac{+\left( -b+\sqrt{{{b}^{2}}+ac} \right)}{a}\text{ and }x=\dfrac{-\left( -b+\sqrt{{{b}^{2}}+ac} \right)}{a}\]

So, the correct answer is “Option B”.

Note: The possibility of mistake in this question can be at the point where \[\left| x \right|=\dfrac{-b-\sqrt{{{b}^{2}}+ac}}{a}\] this value cannot be assumed to be correct. This would be wrong because LHS is positive and RHS is negative is a contradiction. Also, if this step is considered correct then, the possible numbers of real roots would be 4 which is the wrong option.