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

- 0
- 2
- 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±√4b2−4ac2a

Case 1:

b2−4ac<0

⇒ No real solution

Case 2:

b2−4ac>0

4b2>4b2−4ac⇒2b>√4b2−4ac⇒−2b+√4b2−4ac<0⇒−2b+√4b2−4ac2a<0

And −2b−√4b2−4ac2a<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

x2≥0, |x|≥0

Given a,b,c>0

⇒L.H.S.>0 but R.H.S.=0

Hence, the given equation has no real solution.

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±√4b2−4ac2a

Case 1:

b2−4ac<0

⇒ No real solution

Case 2:

b2−4ac>0

4b2>4b2−4ac⇒2b>√4b2−4ac⇒−2b+√4b2−4ac<0⇒−2b+√4b2−4ac2a<0

And −2b−√4b2−4ac2a<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

x2≥0, |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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