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

- no roots
- $2$ real roots
- $3$ real roots
- $4$ real roots

### 5 Comments

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.

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.”

## 6 Answers

### 10 Comments

@ 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. *b*^{2} −4*ac *< 0 There are no real roots.

2. *b*^{2} −4*ac* = 0 There is one real root.

3. *b*^{2} −4*ac *> 0 There are two real roots.

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

@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**

$f(x) = 2x^2 + 2|x| - 3$

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.

Arjun 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.

$ax^{2} + bx + c$, 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).

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.