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Question

If a,b,c are three distinct positive real numbers, then the number of real root(s) of ax2+2b|x|+c=0 is
  1. 0
  2. 2
  3. 4
  4. 1


Solution

The correct option is A 0
The given equation can be rewritten as 
a|x|2+2b|x|+c=0
We know that a,b,c>0 and all are distinct.
Using quadratic formula 
|x|=2b±4b24ac2a

Case 1:
b24ac<0
No real solution
Case 2:
b24ac>0
4b2>4b24ac2b>4b24ac2b+4b24ac<02b+4b24ac2a<0
And 2b4b24ac2a<0

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

Alternate Solution :
ax2+2b|x|+c=0
x20, |x|0
Given a,b,c>0 
L.H.S.>0 but R.H.S.=0
Hence, the given equation has no real solution.

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