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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|x|$$+$$c$$=$$0$ has

1. no roots
2. $2$ real roots
3. $3$ real roots
4. $4$ real roots
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Answer could be easily obtained via graph.
+1

Let the positive roots be $m$ and $n.$ Now, $-m$ and $-n$ will also satisfy the equation $ax^2+b|x|+c=0$ and hence we have $4$ real roots.
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what if n = m ?
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ya, that just clicked  ( x - 2 )2 = x2  - 4x + 4  = 0..

In this case Option b could be true.. :)

but here b != 0

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Sorry, that was plain wrong. Condition for common root is $b^2 = 4ac$.
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@arjun sir

can u explain it more
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Arjun Sir, please comment on this doubt:

A quadratic function is graphically represented by a parabola with vertex located at the origin, below the x-axis, or above the x-axis. Therefore, a quadratic function may have one, two, or zero roots.

The given equation is quadratic and so depending on discriminant we have

1. b2 −4ac < 0 There are no real roots.

2. b2 −4ac = 0 There is one real root.

3. b2 −4ac > 0 There are two real roots.

+2

@manoj

you are right...

but this is not simply a quadratic equation.

if we consider it as quadratic equation considering |x|  as x, you will get two roots which are real (given in the question )

just because of  x^2 and |x|

negative values of above roots(which you have got) also satisfy the above equation

hence total 4 roots

+1 vote
The ans is D  .
+1 vote
The second equation has both +ve and -ve values of roots of first equation as root.

so it will have 4 roots
+1 vote

That I understood

ax2+b|x|+c=0 as x<0 =>  ax2-bx+c=0  ( Two Roots)

x>=0 =>  ax2+bx+c=0 (Two Roots)

=4 Roots