- #1
priscilla98
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Homework Statement
1. x-2|x| < 3
Homework Equations
The Attempt at a Solution
Okay, I attempted this equation below but I don't know if I'm right.
x-2|x|- 3 < 0
x^2 - 2x - 3 < 0
(x - 3) (x + 1) < 0
x = 3 x = -1
priscilla98 said:
x-2|x|- 3 < 0
x^2 - 2x - 3 < 0
(x - 3) (x + 1) < 0
x = 3 x = -1
priscilla98 said:x-2|x|- 3 < 0
x^2 - 2x - 3 < 0
epenguin said:There is just no justification for the step from the first inequality to the second.
Your last line, an equality is unlikely to be a correct answer to this question about an inequality.
But presuming what you mean is at that at these points it changes from < to > or vice versa, even if finding an answer is difficult, checking whether it is the right answer - whether your starting inequalities change at those points you gave - is not and you should not need to ask us.
This is really a very simple question as long as you get straight what it means, particularly what |x| means.
I suggest it is just slightly more obvious if you rewrite the problem in the form
x - 3 < 2|x|
and draw a graph of the two sides.
priscilla98 said:The original problem is
x-2|x|< 3
priscilla98 said:Wait, would you have to get the LCD? But I'm confused, usually when I see fractions, I use the LCD
epenguin said:OK you are trying it out, but you are very handicapped if you don't know what the question means. Which you don't if you don't know what |x| means (where x is any real number).
It just means the quantity x in absolute amount, i.e. always a non-negative number, i.e. |x| = x when x > 0 or x = 0, |x| = -x when x < 0.
Problems involving |x| are usually solved by treating the two cases separately. Edit: as HOI has also said.
(Can I ask has anything I said reminded you of anything said in class? - because with some of the problems I have seen brought here I wonder if students had no notes or no textbooks to look up things.)
HallsofIvy said:The simplest way to do this problem, and any problems involving absolute value, it to break it into cases:
case I: [itex]x\ge 0[/itex] so |x|= x.
The inequality is just x- 2x= -x< 3. Then x> -3 but since we are requiring that [itex]x\ge 0[/math] that says the inequality is true for all [itex]x\ge 0[/itex].
case II: x< 0 so |x|= -x.
The inequality is x+ 2x= 3x< 3 so that x< 1. Since we are requiring that x< 0, that says the inequality is true for all x< 0.
Putting those together, x- 2|x|< 3 is true for all x!
Rational inequalities are inequalities that contain a rational expression. A rational expression is a fraction where the numerator and denominator are polynomials. These types of inequalities often involve finding the values of the variable that make the inequality true.
To graph a rational inequality, first solve the inequality for the variable. Then, plot the solutions on a number line with an open circle for < or > and a closed circle for ≤ or ≥. Finally, shade the area between the solutions to show all the possible values that satisfy the inequality.
To solve rational inequalities algebraically, first find the critical values by setting the numerator and denominator to zero and solve for the variable. Then, plot the critical values on a number line and test a value from each interval to determine if it satisfies the inequality. The intervals that satisfy the inequality will be the solution.
Some common mistakes when solving rational inequalities include forgetting to flip the inequality sign when multiplying or dividing by a negative number, incorrectly factoring the rational expression, and not checking the critical values to ensure they are valid solutions.
Rational inequalities are used in real life to solve problems involving rates, proportions, and budgets. For example, a business owner may use rational inequalities to determine the minimum number of products they need to sell to make a profit. Engineers also use rational inequalities to model and solve real-world problems, such as determining the maximum weight a bridge can hold or the minimum amount of material needed to build a structure.