# 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.

1. there is only one root
2. sum of the roots is + 1
3. sum of the roots is zero
4. the product of the roots is +4

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 ?!

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I think the point of the question is to notice the equation is quadratic in $|x|$ rather than use cases. – Vandermonde Oct 21 '10 at 0:24
The equation is short enough to post in the title of the question. – T.. Oct 21 '10 at 2:21
@T..: Corrected :) – Tretwick Marain Oct 21 '10 at 2:30
Just to be perfectly clear: the minus sign in $-x$ doesn't mean that $-x$ is negative. It means that $-x=0-x$, so it is negative if $x$ is positive and vice versa. I had a teacher who insisted on reading $-x$ as "minus x" rather than "negative x" for this very reason. – Paul VanKoughnett Oct 21 '10 at 2:38
@Paul, tortuous, but I perfectly understand your teacher's insistence. :) I have to say there are still a lot of people who have a hard time grasping "the negative of a negative is positive", both in mathematics and in language. – J. M. Oct 21 '10 at 2:50

$\rm\ 0\ =\ x^2 + |x| -6\ =\ (|x| - 2)\:(|x| + 3)\ \Rightarrow\ |x| = 2\ \Rightarrow\ x = \pm2$

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$f(x)=x^2+|x|-6$ 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.

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+1. But you also use the fact that it has finitely many zeros. :) – Jonas Meyer Oct 21 '10 at 5:03
@Jonas Meyer: I suppose I do assume that the sum of the zeros exists (which I don't think requires finitely many zeros). – Isaac Oct 21 '10 at 5:49
OK, perhaps you assume at most countably many zeros with 0 as the only possible limit point. – Jonas Meyer Oct 21 '10 at 6:07
I think it's pretty clear that there are finitely many (real) zeroes; in fact there at most 4, since x^2 + |x| - 6 = 0 implies x^2 = (6 - x^2)^2. – Qiaochu Yuan Oct 21 '10 at 16:46
Yes, or because $f$ is a polynomial function on $(0,\infty)$ and on $(-\infty,0)$. I was just having fun, but also pointing out that technically something about the function must be used beside the fact that it is even. – Jonas Meyer Oct 21 '10 at 17:35

If x<0, then -x>0, which means that the modulus is indeed positive.

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Then which will be the correct option here ? I think (2) since the sum of the two roots is 1, isn't ? – Tretwick Marain Oct 21 '10 at 0:02
The point of the answer is that you can only accept one of the two roots of (each case of) the quadratic due to the absolute value sign. So there is one root of 2, coming from the first case where x>0 is assumed. The root of -3 was rejected as inconsistent with the assumption. In the second case, x<0 is assumed, and only the root of -2 is acceptable. Interpreting what jake said, if x=-2, |x|=2, which means |x|=-x. Adding the two roots together gives zero. – Ross Millikan Oct 21 '10 at 0:11
@Ross Millikan :Aha! that's a nice explanation Ross :) Really appreciated :) – Tretwick Marain Oct 21 '10 at 0:15

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 $y = x + |x|$

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Well,nice explanation:) How did you add the graph here ? ;) – Tretwick Marain Oct 21 '10 at 0:29
@Tretwick: Heh, I just went to Wolfram Alpha and plotted it there, and uploaded the image to imageshack. – Justin L. Oct 21 '10 at 0:33
I also noted that I completely missed the point of the question. – Justin L. Oct 21 '10 at 0:36