Exploring the quadratic equation $x^2 + \lvert x\rvert - 6 = 0$ This question and the described solution are copied from a test-paper :
For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.


*

*there is only one root

*sum of the roots is + 1

*sum of the roots is zero

*the product of the roots is +4


Answer : (3)
Answer Explanation : 
If x > 0 |x| = x.
Given equation will be $x^2 + x - 6 = 0$⇒ x = 2,- 3 ⇒ x = 2
If x < 0 |x| = - x.
Given equation will b e  $x^2$ - x - 6 = 0 ⇒ x = -2, 3 ⇒x = - 2
Sum of roots is 2 - 2 = 0
Now I have a doubt on the statment "If x < 0 |x| = - x." I think modulus means that |x| is always positive ?! Also I can see that (2) seems to be the correct option isn't ?!
Please post your views.
 A: $f(x)=x^2+|x|-6$ is an even function—that is, $f(x)=f(-x)$ for all $x$, or the graph of $y=f(x)$ is symmetric over the y-axis—so if $f(c)=0$ then $f(-c)=0$, so the sum of the zeros of $f$ must be 0.
A: $\rm\ 0\ =\ x^2 + |x| -6\ =\ (|x| - 2)\:(|x| + 3)\ \Rightarrow\ |x| = 2\ \Rightarrow\ x = \pm2 $
A: If x<0, then -x>0, which means that the modulus is indeed positive.
A: The statement is saying:
$|x| = -x$ (for $x < 0$)
and divide both sides by $x$ to give us:
$\frac{|x|}{x} = -1$
Let us test a few values to make sure this holds.
$x = -3$ and $|x| = 3$:
$\frac{|-3|}{-3} = -1$
$\frac{3}{-3} = -1$
$-1 = -1$
True.
$x = -5$ and $|x| = 5$
$\frac{5}{-5} = -1$
$-1 = -1$
True.
We can verify this for all $x < 0$.
If you don't believe me, here is a plot of $y = \frac{|x|}{x}$.


Alternative:
If we define
-a is the number such that a + (-a) = 0

ie,
if a = 3, -a = -3, because $3 + (-3) = 0$
if a = -2, -a = 2, because $-2 + 2 = 0$
We can call 3 and -3 "opposites".
The opposite of -2 is 2.
The opposite of 5 is -5.
Then we can translate the original statement then as:
If x < 0, |x| is the opposite of x

Let's test this out.
$x = -4$
$|x| = 4$
$-4 + 4 = 0$
$|-4|$ (or, 4) is indeed the opposite of $-4$.
More generally, the statement says that, if $x < 0$,
$x + |x| = 0$
Which we can verify as true for all values $x < 0$
If you don't believe me, you can look at the plot of $y = x + |x|$

