# Quadratic inequalities with absolute values

• Callmelucky
In summary: If a problem is not well posed, then there may not be any solutions at all. This is not the case here, but some of the inequalities might be redundant.
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.

Last edited:
Callmelucky said:
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.

Callmelucky
Callmelucky said:
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
Mark44 said:
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.

Mark44 said:
##|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.

Callmelucky said:
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.

Callmelucky said:
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.

You could even express the solution set as $$\begin{split} x &\in \left((-\infty, -4) \cap (-5, - 1)\right) \cup \left((-2, \infty) \cap [-2 - \sqrt{3}, -2 + \sqrt{3}]\right) \\ &= [-5, - 4) \cup (-2, -2 + \sqrt{3}] \end{split}$$ 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.

## 1. What is a quadratic inequality with absolute values?

A quadratic inequality with absolute values is an inequality that involves a quadratic expression with an absolute value sign. This means that the inequality has two possible solutions, one with the absolute value being positive and one with the absolute value being negative.

## 2. How do you solve quadratic inequalities with absolute values?

To solve a quadratic inequality with absolute values, you must first isolate the absolute value expression. Then, you can set up two separate inequalities, one with the absolute value being positive and one with the absolute value being negative. Solve each inequality separately and combine the solutions to find the final solution set.

## 3. What are the key steps in solving quadratic inequalities with absolute values?

The key steps in solving quadratic inequalities with absolute values are: isolating the absolute value expression, setting up two separate inequalities, solving each inequality separately, and combining the solutions to find the final solution set.

## 4. How do you determine the direction of the inequality in a quadratic inequality with absolute values?

The direction of the inequality in a quadratic inequality with absolute values is determined by the sign of the coefficient of the quadratic term. If the coefficient is positive, the parabola opens upwards and the inequality will be greater than or equal to. If the coefficient is negative, the parabola opens downwards and the inequality will be less than or equal to.

## 5. Can quadratic inequalities with absolute values have more than two solutions?

Yes, quadratic inequalities with absolute values can have more than two solutions. This is because the absolute value expression can have two possible solutions, one positive and one negative, which can result in multiple solutions for the overall inequality.

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