edited by
8,763 views
29 votes
29 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
edited by

6 Answers

Best answer
42 votes
42 votes
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.

Correct Answer: $D$
edited by
23 votes
23 votes

$ax^{2} + bx + c$, for roots to be real & positive,
Discriminant: $b^{2}$ − 4ac > 0

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

And,

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

Therefore, correct answer is option (D).

edited by
7 votes
7 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

               

3 votes
3 votes

Since mod(x) is given, we have to evaluate 2 conditions which are (1) +x &  (2) -x

For the equation $ax^{2} + bx + c = 0$ , we get 2 real roots.
For the equation $ax^{2} - bx + c = 0$ , we get 2 real roots.
So, totally, we get 2 + 2 = 4 real roots.

Ans.: (D) 4 real roots

Answer:

Related questions

31 votes
31 votes
6 answers
1
gatecse asked Sep 15, 2014
8,643 views
When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created wit...
20 votes
20 votes
3 answers
2
gatecse asked Sep 15, 2014
4,842 views
In a survey, $300$ respondents were asked whether they own a vehicle or not. If yes, they were further asked to mention whether they own a car or scooter or both. Their r...
21 votes
21 votes
3 answers
3