Solving an absolute value quadratic inequality

[sylent33]

Homework Statement

Find all the real solutions

Relevant Equations

Equation solving

Im a having trouble understanding how this exactly works.

$$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.

1. $x^2-4 >= 0$ and 2. $x^2-4 < 0$

Out of that 1. $x >=\pm 2$ and 2. $x<\pm 2$

For the first case the term $x^2 -4$ stays positive and for the solution I get 0 < 6 which is a true statement for all real number so the solution should be R. Now the part that I dont get. We have this boundary set by the fact that $x^2 -4$ is positve and it says that x has to be great or equal than -2 and +2. So it should be all the real numbers from 2 so $L_1 = [2,+\inf]$ For the second part case

$$ -x^2+4 < x^2+2 $$ I get $x>\pm 1$ and using the bondary I have for case 2 that it has to be less than -2 and +2 I get $L_2 = (-2,-1) \cup (1,2)$

Now for the final solution I put $L = (-2,-1) \cup [2, \infty)$ That is not correct,the solution sheet says it should be $$ L =(-\infty,1) \cup (1, \infty) $$

I dont understand how they got here so I checked what they got for L1 and L2. L2 matches but for L1 they have $L_1 = R \backslash {(2,-2)}$

The way I am interpreting this is,all real numbers except 2 and -2. I do not understand how this can be their solution. It is great EQUAL shouldnt 2 and -2 be included,Also we are looking for solutions GREATER than -2 and 2 so I though since 2 is greater than -2 I only need to go from 2 to infinity. Why do I have to include minus infinity?

 

[pasmith]

Homework Statement:: Find all the real solutions
Relevant Equations:: Equation solving

Im a having trouble understanding how this exactly works.

$$ |x^2 - 4| < |x^2+2| $$

Note that the inequality only involves $x^2$. So if it is true for $x$, it is also true for $-x$. If your solution set does not have that property then you have made an error.

So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.

1. $x^2-4 >= 0$ and 2. $x^2-4 < 0$

Out of that 1. $x >=\pm 2$ and 2. $x<\pm 2$

Please don't write expressions like these; they are meaningless. The first is "x <= -2 or x >= 2" and the second is "-2 < x < 2".

For the first case the term $x^2 -4$ stays positive and for the solution I get 0 < 6 which is a true statement for all real number so the solution should be R. Now the part that I dont get. We have this boundary set by the fact that $x^2 -4$ is positve and it says that x has to be great or equal than -2 and +2. So it should be all the real numbers from 2 so $L_1 = [2,+\inf]$

You have, I think, confused yourself by writing imprecise statements like "greater than or equal to -2 and +2". As I pointed out above, what you actually have here is $x \leq -2$ or $x \geq 2$, so $$L_1 = (-\infty, -2] \cup [2, \infty) = \mathbb{R} \setminus (-2,2).$$

For the second part case

$$ -x^2+4 < x^2+2 $$ I get $x>\pm 1$ and using the bondary I have for case 2 that it has to be less than -2 and +2 I get $L_2 = (-2,-1) \cup (1,2)$

This is correct, but don't write $x > \pm 1$.

Now for the final solution I put $L = (-2,-1) \cup [2, \infty)$

What about $(1,2)$? You correctly had that as part of $L_2$, so why did you drop it here?

There is however an easier way to solve this inequality: it has the geometric interpretation that $x^2$ is closer to 4 than it is to -2. The mid-point is 1, so $x^2 > 1$. Hence either $x < -1$ or $x > 1$.

 

[Steve4Physics]

1. $x^2-4 >= 0$ and 2. $x^2-4 < 0$

Out of that 1. $x >=\pm 2$ and 2. $x<\pm 2$

That's wrong/confusing. @pasmith beat me to it, but since I've already written it:

For 1: $x^2 - 4 \ge 0$ means $x \ge 2$ or $x \le - 2$.
For 2: $x^2 - 4 \lt 0$ means $x \gt -2$ and $x \lt 2$ (i.e. $-2 \lt x \lt 2)$.

By the way, the LaTex symbol for $\ge$ is simply \ge for example.

 

[Mark44]

here it is even simpler since the right part of the inequality is always positive

In fact, the expression on the right side of the inequality, $|x^2 + 2|$ is always greater than or equal to 2, for any real number x.

 

[sylent33]

Ah I see now,writing out like that makes it really simple. I did not know how to deal with that.

Thanks a lot for the help

PS: I plotted both of these functions and you can clearly see at which value they are the same,and that the solution in the book is correct.

 

[epenguin]

I came back to this thread because I have noticed recently several people making heavy weather of this same kind of problem with absolute values. but you have done exactly what I was going to say, plotted the function. Or just sketch them.

To plot | f(x) | you just plot f(x) and then all that is below the X axis is reflected above it. Then the answer becomes obvious (or what do you have to do in the case of detailed calculation does).

Maybe you are being examined for more, for the formal setting out of argument. But it will be helpful to know the answer anyway. I questio the didactic value of such exercises, I don't think one encounters them much in maths applications or that it wouldn't be in the same way obvious out how to proceed if one dId, so I don't think it's a good idea to turn these things into a chore.

 

[pasmith]

Maybe you are being examined for more, for the formal setting out of argument. But it will be helpful to know the answer anyway. I questio the didactic value of such exercises, I don't think one encounters them much in maths applications or that it wouldn't be in the same way obvious out how to proceed if one dId, so I don't think it's a good idea to turn these things into a chore.

These things crop up in the context of numerical schemes for solving partial differential equations, where the condition that the scheme be stable often reduces to the condition that the timestep $\Delta t > 0$ is such that $$\left|\frac{1 + \Delta t P(\lambda)}{1 - \Delta tQ(\lambda)}\right| < 1$$ for every $\lambda$ in some given subset of $\mathbb{C}$, where the polynomials $P$ and $Q$ are determined by the details of the scheme in question.

 

[epenguin]

These things crop up in the context of numerical schemes for solving partial differential equations, ...

Yes, I also did think of an application but do you think there will have been any benefit to have been doing exercises on it to no apparent purpose at the time two or more years previously? More motivating to do it when you need it, purpose and meaning then more evident.?

 

[Mark44]

I came back to this thread because I have noticed recently several people making heavy weather of this same kind of problem with absolute values. but you have done exactly what I was going to say, plotted the function.

You get no argument from me that plotting the function is a good idea, but once a student understands how such graphs look in general, he/she should also be able to convert an equation or inequality with absolute values into the corresponding form that doesn't use absolute values.
That is, an inequality such as |f(x)| < a (where a > 0), can be rewritten as -a < f(x) < a, but not as the OP wrote it: $f(x) < \pm a$. And similarly for |f(x)| > a (where a > 0), with the inequalities f(x) < -a or f(x) > a.

Yes, I also did think of an application but do you think there will have been any benefit to have been doing exercises on it to no apparent purpose at the time two or more years previously?

Applications can come much sooner than "two or more years" later from when the topic is first presented. Often, the presentation of absolute values comes within a few months before the first quarter or semester of calculus, when differentiation is discussed; e.g., finding $\frac d{dx}|x|$. Soon after that, the student might be tasked with finding the area beneath curves such as y = |f(x)|.
Following that, the student likely will be learning about infinite series. Several tests involve finding the limit of ratios of the form $\frac{|a_{n+1}|}{a_n}}$. Inequalities such as these don't usually lend themselves to being graphed very easily. Another application of absolute values is in estimating the error in approximating a function's value by using a fixed number of terms in the functions Maclaurin or Taylor series.

 

[Mark44]

A minor problem with the LaTeX.

Thanks! Fixed in the original post and in the quoted text.