# Solving an absolute value quadratic inequality

• sylent33
In summary: If you ignore this, you should have ##x \ge 2## and ##-2 \ge x##, which is a contradiction.Here's a way to avoid the confusion: start with the obvious statement that ##|a| < |b|## is the same as ##-|b| < a < |b|##. Then we have ##-|x^2 + 2| < x^2 - 4 < |x^2 + 2|##. We don't need to break it up into cases, because we can see that the left inequality implies that x^2 + 2 > 0, so x >
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?

sylent33 said:
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 $x > 1$.

MatinSAR, sylent33 and Steve4Physics
sylent33 said:
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.

MatinSAR and sylent33
sylent33 said:
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.

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.

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.

malawi_glenn and sylent33
epenguin said:
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.

pasmith said:
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.?

epenguin said:
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.

epenguin said:
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.

SammyS said:
A minor problem with the LaTeX.
Thanks! Fixed in the original post and in the quoted text.

## 1. What is an absolute value quadratic inequality?

An absolute value quadratic inequality is an inequality that contains a quadratic expression with an absolute value. It can be written in the form |ax^2 + bx + c| < d, where a, b, c, and d are real numbers and x is the variable.

## 2. How do I solve an absolute value quadratic inequality?

To solve an absolute value quadratic inequality, you first need to isolate the absolute value expression on one side of the inequality. Then, you can split the inequality into two separate inequalities, one with a positive sign and one with a negative sign. Solve each inequality separately and then combine the solutions to get the final solution.

## 3. Can an absolute value quadratic inequality have multiple solutions?

Yes, an absolute value quadratic inequality can have multiple solutions. This is because the absolute value function can have two possible values for a given input, one positive and one negative. Therefore, when solving an absolute value quadratic inequality, you may end up with two separate solutions.

## 4. How do I graph an absolute value quadratic inequality?

To graph an absolute value quadratic inequality, you can first graph the corresponding quadratic function without the absolute value. Then, you can use the behavior of the absolute value function to determine which parts of the graph to reflect over the x-axis. The resulting graph will be a V-shaped curve or two separate curves, depending on the number of solutions.

## 5. Can I use the quadratic formula to solve an absolute value quadratic inequality?

Yes, you can use the quadratic formula to solve an absolute value quadratic inequality. However, you will need to first isolate the absolute value expression and then solve the resulting quadratic equation using the quadratic formula. This method may be more time-consuming compared to splitting the inequality into two separate inequalities.

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