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+19 votes

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
asked in Numerical Ability by Boss (18.2k points)
edited by | 2.7k views
Answer could be easily obtained via graph.
can u upload sample piece?

5 Answers

+28 votes
Best answer
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.
answered by Veteran (395k points)
selected by
what if n = m ?

ya, that just clicked  ( x - 2 )2 = x2  - 4x + 4  = 0..

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

but here b != 0

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

can u explain it more

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.



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

+4 votes

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


answered by Active (4.7k points)
+2 votes
$ax^{2} + bx + c$, for roots to be real & +ve $b^{2}$ − 4ac > 0
This will have 2 real positive roots.

$ax^{2} + b|x| + c$ = 0
This can be written as
i) $ax^{2} + bx + c$ = 0   (x>=0)
Discriminant = $b^{2}$ − 4ac > 0. --> Roots are real and positive


ii) $ax^{2} - bx + c$​​​​​​​ = 0   (x<0)
$(-b)^{2}$ − 4ac ⇒ $b^{2}$ − 4ac is also >0. --> Roots are real and positive
So, we will have 4 real roots for $ax^{2} + b|x| + c$​​​​​​​ = 0.

Therefore, correct answer is option (D).
answered by Active (1.7k points)
+1 vote
The ans is D  .
answered by Active (1.5k points)
+1 vote
The second equation has both +ve and -ve values of roots of first equation as root.

so it will have 4 roots
answered by Loyal (9.3k points)

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