Solving for ##x## for a given inequality

In summary, the problem is to solve the inequality ##\dfrac{x}{x+2}\le \dfrac{1}{|x|}##, with the restrictions that ##x\ne 0, -2## and ##x>0##. The solution involves considering two cases, one where ##x>0## and the other where ##x<0##. After simplifying and analyzing each case, the final solution is ##\boxed{-2<x<-1\;\;\;\text{OR}\;\;\; 0<x<2}##.
  • #1
brotherbobby
651
156
Homework Statement
Solve : ##\mathbf{\dfrac{x}{x+2}\le \dfrac{1}{|x|}}##
Relevant Equations
(1) By definition, ##|x| = x## if ##x\ge 0## and ##|x| = -x## if ##x \le 0##.
(2) ##\frac{1}{x}## is not defined for ##x = 0##.
(3) If ##|x| > a\Rightarrow x>a\; \text{or}\; x<-a##
1656848675312.png
Problem Statement :
I copy and paste the problem from the text to the right.

Attempt (mine) : Given the inequality ##\dfrac{x}{x+2}\le \dfrac{1}{|x|}##. We see immediately that ##x\ne 0, -2##. At the same time, since ##|x|\ge 0\Rightarrow \frac{x}{x+2}\ge 0##.

Now if ##\frac{x}{x+2}\le \frac{1}{|x|}\Rightarrow \frac{1}{|x|}\ge \frac{x}{x+2}##, then we have either ##\frac{1}{x}\ge \frac{x}{x+2}## or ##\frac{1}{x}\le -\frac{x}{x+2}##.

1. The first inequality above reduces to ##\frac{1}{x}-\frac{x}{x+2}\ge 0\Rightarrow \frac{x+2-x^2}{x(x+2)}\ge 0\Rightarrow \frac{x^2-x-2}{x(x+2)}\le 0\Rightarrow \frac{(x-2)(x+1)}{x(x+2)}\le 0## which lead to the answers : ##\boxed{-2<x<-1\; \text{OR}\; 0<x<2}## (note the equality condition is forbidden due to considerations right at the beginning). This can be found by the number line
1656849968462.png
pneumonic shown to the right or any other way you are used to with inequalities.

I attach the pictorial pneumonic to the right for those who are used to it. 2. The second inquality above reduces to ##\frac{1}{x}+\frac{x}{x+2}\le 0\Rightarrow \frac{x^2+x+2}{x(x+2}\le 0##. Now we can show using the method of "squaring a quadratic equation" that the numerator ##x^2+x+2>0 \;\;\forall x##. Thus focus falls on the numerator, which yields ##\boxed{-2<x<0}##.

Putting both cases 1 and 2 together and caring for overlaps, my solution for the proble reads : ##\boxed{-2<x<-1\;\;\;\text{OR}\;\;\; 0<x<2}## .

Solution (Text) : I could neither follow, nor agree, with the solution in the text. I copy and paste it below.

1656850738316.png


I have underlined what I believe to be the solutions. So ##x=0?!##. Clearly that is not correct. However, the text is right when I take samples from its solutions that are less than zero, for instance ##x = -1, -0.5## etc., solutions which I did not get. Surely my method is mistaken.

A hint or suggestion would be welcome.
 
Physics news on Phys.org
  • #2
Your case 1 is when ##x>0## thus the solution ##-2<x<-1## must be disregarded as out of the allowed limit (which is ##x>0##).
 
  • Like
Likes FactChecker
  • #3
In addition to @Delta2's point, you have combined your two 'ORed' cases incorrectly. You should have included all of the -2<x<0 results of the second case.
 
  • #4
brotherbobby said:
I attach the pictorial pneumonic
"Pneumonic" isn't the word you want here. That would be "mnemonic."
 
  • Love
  • Haha
Likes nuuskur and Delta2

Related to Solving for ##x## for a given inequality

1. How do I solve for ##x## in an inequality?

To solve for ##x## in an inequality, you need to isolate the variable on one side of the inequality symbol. This can be done by using inverse operations, such as adding, subtracting, multiplying, or dividing both sides of the inequality by the same number. Remember to also apply the same operation to both sides to maintain the balance of the inequality.

2. What is the difference between solving for ##x## in an equation and an inequality?

The main difference between solving for ##x## in an equation and an inequality is that in an equation, the goal is to find a specific value for ##x## that makes the equation true. In an inequality, the goal is to find all possible values of ##x## that make the inequality true. This means that the solution to an inequality is often expressed as a range of values, rather than a single value.

3. How do I know if my solution to an inequality is correct?

To check if your solution to an inequality is correct, you can substitute the value of ##x## into the original inequality and see if it makes the inequality true. If it does, then your solution is correct. You can also graph the inequality and see if the solution falls within the shaded region of the graph.

4. Can I use the same steps to solve all types of inequalities?

While the general steps for solving inequalities are the same, there are some differences in the specific steps depending on the type of inequality. For example, when solving for ##x## in a compound inequality (an inequality with two or more inequality symbols), you need to consider the order of operations and the direction of the inequality symbols. It is important to carefully read and understand the given inequality before attempting to solve it.

5. What happens if the inequality has a variable on both sides?

If the inequality has a variable on both sides, you will need to use the same steps as you would for an equation with variables on both sides. This involves combining like terms and isolating the variable on one side of the inequality symbol. Once you have isolated the variable, you can continue solving for ##x## using the same steps as you would for a regular inequality.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
13
Views
424
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
511
  • Precalculus Mathematics Homework Help
Replies
10
Views
987
  • Precalculus Mathematics Homework Help
Replies
10
Views
919
  • Precalculus Mathematics Homework Help
Replies
5
Views
796
  • Precalculus Mathematics Homework Help
Replies
7
Views
940
  • Precalculus Mathematics Homework Help
Replies
5
Views
262
  • Precalculus Mathematics Homework Help
Replies
15
Views
755
Back
Top