Thread: Solving of inequalities involving modulus

1. Solving of inequalities involving modulus

"Solve the inequality 1/(x-a) < 4|x-a|, where a is a positive constant, leaving your answer in terms of a." The answers are x <a and x > a + 1/2.

What I've got:

I know the above inequality gives 4(x - a) < -1/(x - a) or 4(x - a) > 1/(x - a).

The inequality on the left gives x < a , while the inequality on the right gives a - 1/2 < x < a or x > a + 1/2. Taking the union of the 2 solutions gives x < a or x > a + 1/2.

I can get x < a but I can't get a - 1/2 < x < a or x > a + 1/2. Can anyone help me out here? thanks! Please detail the steps, I've been working on it for a long time but can't get the answer

2. The given inequality 1/(x-a) < 4l x-a l

corresponding equation 4l x-a l = 1/(x-a)

gives 4(x-a) = -1 / (x-a) or 4(x-a) = 1/ (x-a)

1st gives no result for 2nd (x-a)^2 = 1/4

x-a = 1/2 or x-a = -1/2

x = a + 1/2 and x = a - 1/2 with x = a as free boundry

on number line we get the required result

3. you disregarded the inequality sign, and for 4(x - a) < -1/(x - a), i eventually got ((2x - 2a)^2 +1)/(a-x) > 0, then 1/(a-x) > 0, then a-x >0 and x < a. But I don't know how to get x-a = 1/2 or x-a = -1/2.

can you detail, step by step (or at least the first few steps) how you got x-a = 1/2 or x-a = -1/2? thanks ((: sorry i don't really know how to get there )): I've tried for really long i think my method is wrong that's why i want the steps

4. Originally Posted by thesocialnetwork
"Solve the inequality 1/(x-a) < 4|x-a|, where a is a positive constant, leaving your answer in terms of a." The answers are x <a and x > a + 1/2.

What I've got:

I know the above inequality gives 4(x - a) < -1/(x - a) or 4(x - a) > 1/(x - a).

The inequality on the left gives x < a , while the inequality on the right gives a - 1/2 < x < a or x > a + 1/2. Taking the union of the 2 solutions gives x < a or x > a + 1/2.

I can get x < a but I can't get a - 1/2 < x < a or x > a + 1/2. Can anyone help me out here? thanks! Please detail the steps, I've been working on it for a long time but can't get the answer
You must take care when solving rational inequalities. Clearing fractions CAN and usually will lead to incorrect results.

Starting with your equation you have that

$4(x-a) < -\frac{1}{x-a} \iff \frac{4(x-a)^2}{x-a}< -\frac{1}{x-a} \iff \frac{4(x-a)^2}{x-a}+\frac{1}{x-a} < 0 \iff \frac{4(x-a)^2+1}{x-a} < 0$

Since the left hand side is always positive the equation has not solutions. Now for the 2nd equation you get

$4(x-a) > \frac{1}{x-a} \iff \frac{4(x-a)^2}{x-a} > \frac{1}{x-a} \iff \frac{4(x-a)^2}{x-a}-\frac{1}{x-a} > 0 \iff \frac{[2(x-a)+1][2(x-a)-1]}{x-a} < 0$

This gives the three critical points $x=a,x=a \pm \frac{1}{2}$

5. Thanks! Great presentation, you've been of much help 8D

6. It is one of the methods used to solve inequalities , using corresponding equation
First we find the roots of corresponding equation along with free boundries ( the values of x for which equation is undefined )
Next we check values of x within the above boundries in inequality for true or false.