GATE CSE 2014 Set 1 | Question: GA-5
in Quantitative Aptitude edited by
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26 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
in Quantitative Aptitude edited by
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5 Comments

Answer could be easily obtained via graph.
0
can u upload sample piece?
2
edited by
None of the answer is correct for all possible quadratic equations. Here is a counter example.

For $x^{2} - 2x + 1 = 0$, the only root is $1$ and it satisfies the criteria of being real & positive. Also the other criteria of  $a, b, c$ being real numbers is satisfied as well.

Now, the equation $x^2 -2|x| + 1 = 0$ can be defined as two equations $x^{2} - 2x + 1 = 0$, $x \geq 0$ and $x^{2} + 2x + 1 = 0$, $x < 0$ which only have a total of 2 roots which are $1$ and $-1$ respectively and not 4 roots.

All quadratic equations where the discriminant is zero and $b$ & $a$ are of same sign will only give 2 roots for such transformation.
0

correct statement should be: $x^2- 2|x|+1 = 0$ has $2$ $\textbf{distinct}$ real roots in which root $”1”$ has multiplicity = $2$ and root $“-1”$ also has multiplicity as $2$ because as you have written, equation $x^2-2x+1=0$ for $x \geq 0$ has $2$ $\textbf{equal}$ real roots (discriminant=0) which has value = $“1”$ and $x^2+2x+1=0$ for $x < 0$ also has $2$ $\textbf{equal}$ real roots which has value = $“-1”$ So, total real roots are $4$.

This is based on the following statement from the Fundamental theorem of algebra:

“every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.” 

0

@ You are right. Thanks.

0

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6 Answers

37 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.

Correct Answer: $D$
edited by
by

10 Comments

what if n = m ?
1

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

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

but here b != 0

0
Sorry, that was plain wrong. Condition for common root is $b^2 = 4ac$.
5
@arjun sir

can u explain it more
0

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.

http://www.biology.arizona.edu/biomath/tutorials/Quadratic/Roots.html

1

@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

2

Can someone tell me why the equation when "graphed' gives only 2 roots? In the first image only 1 equation (red) is highlighted. In second image, 2 equations are highlighted. 

0

@commenter commenter

$f(x) = 2x^2 + 2|x| -  3$

for $x \geq 0$, $f(x) = 2x^2 +2x - 3$

and for $x < 0$, $f(x) = 2x^2 -2x - 3$

Now, for $x \geq 0$, equation $f(x) = 0$ has only $1$ positive real root.

and for $x < 0$, equation $f(x) = 0$ has only $1$ negative real root.

So, by considering both portions of $X$-axis i.e. $x \geq 0$ and $x<0$ , equation $2x^2 + 2|x| - 3 = 0$ has $2$ real roots. 

1

Okay. So, I forgot that we need to consider only positive roots of the equation. If we take equation with positive roots then it has 4 roots (provided they are distinct). 

0

 sir, here for x>0, the equation has real roots but for x<0 the equation can have either both real or both complex roots. Either way the total toots of equation will be 4.

0
15 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
6 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

               

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

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

0 votes
None of the answer is correct for all possible quadratic equations. Here is a counter example.

For $x^{2} - 2x + 1 = 0$, the only root is $1$ and it satisfies the criteria of being real & positive. Also the other criteria of  $a, b, c$ being real numbers is satisfied as well.

But even then the two equations $x^{2} - 2x + 1 = 0$, $x \geq 0$ and $x^{2} + 2x + 1 = 0$, $x < 0$ only have a total of 2 roots which are $1$ and $-1$ and not 4 roots.

All quadratic equations where the determinant is zero and $b$ is positive will only give 2 roots for such transformation.
Answer:

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