Quadratic inequalities with absolute values

[Callmelucky]

Homework Statement

What is the right way to write solutions?

Relevant Equations

$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

I was given a problem to solve that goes like this $\frac{3}{|x+3|-1}\geq |x+2|$ . I got the correct solution for all possible cases and here they are; for $|x+3|\geq0$ and $|x+2|\geq$ i got $x\epsilon <-2, -2\sqrt{3} ]$ and for $|x+3|\leq0$ , $|x+2|\leq0$ I got $x\epsilon [-5, -4>$
For other cases there is no possible solutions.
Now, my question is, since these are answers for different scenarios $|x+3|\geq0$ and $|x+2|\geq0$ and $|x+3|\leq0$ and $|x+2|\leq0$, should I leave solutions as answers separately or should I write them as union eg. $x\epsilon [-5, -4> U <-2, -2+\sqrt{3}]$?
Thank you.

 

[Mark44]

Homework Statement:: What is the right way to write solutions?
Relevant Equations:: $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

I was given a problem to solve that goes like this $\frac{3}{|x+3|-1}\geq |x+2|$ . I got the correct solution for all possible cases and here they are; for $|x+3|\geq0$ and $|x+2|\geq$ i got $x\epsilon <-2, -2\sqrt{3} ]$ and for $|x+3|\leq0$ , $|x+2|\leq0$ I got $x\epsilon [-5, -4>$
For other cases there is no possible solutions.
Now, my question is, since these are answers for different scenarios $|x+3|\geq0$ and $|x+2|\geq0$ and $|x+3|\leq0$ and $|x+2|\leq0$, should I leave solutions as answers separately or should I write them as union eg. $x\epsilon [-5, -4> U <-2, -2+\sqrt{3}]$?
Thank you.

You have some overlap in your inequalities. $|x + 3| \ge 0$ but $|x + 3| \le 0$ has only 1 solution (x = -3), and no solutions for $|x + 3| \lt 0$. Same is true for |x + 2|.
Assuming your work is correct, you can write the intervals as a union or with the word "or" between them.

Either of the below would be fine.
$x \in [-5, -4) \cup (-2, -2 + \sqrt 3]$
$-5 \le x \lt -4 \text{ or } -2 \le x \lt -2 + \sqrt 3$

Note that to indicate the inclusion of an endpoint, one common notation is a bracket, but to exclude an endpoint, a parenthesis is often used.

BTW, you probably should use \in rather than \epsilon. They look similar, but the first is easier to type.

 

[Mark44]

I got the correct solution for all possible cases and here they are; for $|x+3|\geq0$ and $|x+2|\geq$

$|x + 3| \ge 0$ by the definition of the absolute value, but |x + 3| can't be equal to 1. This means that $x \ne -2$ and $x \ne -4$. Your solution shows this, but I don't see that you have considered this otherwise.

 

[Callmelucky]

You have some overlap in your inequalities. $|x + 3| \ge 0$ but $|x + 3| \le 0$ has only 1 solution (x = -3), and no solutions for $|x + 3| \lt 0$. Same is true for |x + 2|.
Assuming your work is correct, you can write the intervals as a union or with the word "or" between them.

Either of the below would be fine.
$x \in [-5, -4) \cup (-2, -2 + \sqrt 3]$
$-5 \le x \lt -4 \text{ or } -2 \le x \lt -2 + \sqrt 3$

Note that to indicate the inclusion of an endpoint, one common notation is a bracket, but to exclude an endpoint, a parenthesis is often used.

BTW, you probably should use \in rather than \epsilon. They look similar, but the first is easier to type.

Thank you, but if I write it with the union, wouldn't that mean that for both cases both solutions are valid? Which they are not.
Thanks for the tip.

 

[Callmelucky]

$|x + 3| \ge 0$ by the definition of the absolute value, but |x + 3| can't be equal to 1. This means that $x \ne -2$ and $x \ne -4$. Your solution shows this, but I don't see that you have considered this otherwise.

I haven't. But I know I need to do that. Thanks for letting me know.

 

[Mark44]

Thank you, but if I write it with the union, wouldn't that mean that for both cases both solutions are valid? Which they are not.
Thanks for the tip.

I didn't work through the whole problem, so I don't know which cases are valid or invalid. If you have some cases that turn out to be invalid, then your work along the way should have eliminated those cases from consideration.

The left side of your inequality is $\frac 3 {|x + 3| - 1}$. The connector is $\ge$, which means the fraction must be nonnegative. For this to happen, |x + 3| - 1 must be nonnegative, and |x + 3| can't be equal to 1. Any cases you develop should take these constraints into account.

 

[PeroK]

Thank you, but if I write it with the union, wouldn't that mean that for both cases both solutions are valid? Which they are not.
Thanks for the tip.

Or and union in this context are equivalent.

 

[pasmith]

You could even express the solution set as \begin{split}<br /> x &amp;\in \left((-\infty, -4) \cap (-5, - 1)\right) \cup \left((-2, \infty) \cap [-2 - \sqrt{3}, -2 + \sqrt{3}]\right) \\<br /> &amp;= [-5, - 4) \cup (-2, -2 + \sqrt{3}] \end{split}<br /> since in both cases the solution must (1) lie between the two roots of a quadratic, and (2) satisfy the conditions under which that quadratic was derived.