What Steps Solve Modulus Inequalities in Algebra?

[Sumedh]

Homework Statement

Solve \frac{|x^2-5x+4|}{|x^2-4|}\le1

Homework Equations

The Attempt at a Solution

as

|x^2-4|will be positive always

cross multiply and take 1 to other side of equation
solve by taking LCM
we get
|x^2-5x+4|-(x^2-4)\le0
on solving we get

(x^2-5x+4)-(x^2-4)\le0 and -(x^2-5x+4)-(x^2-4)\le0

the other method I know is to square to remove the modulus function
(x^2-5x+4)^2-(x^2-4)^2\le0


among these which method is correct?
the second method becomes equation of degree 4 i.e.x^4...


please provide hints.

 

[rock.freak667]

The Attempt at a Solution

as

|x^2-4|will be positive always

Actually for x=1, 'x²-4' is negative, but you can use the fact that |a|/|b| = |a/b| iff b≠0.

Just use the fact that |X|<A ⇒ -A<X<A and then just take each inequality separately and take the union of the sets.

 

[ehild]

|x^2-5x+4|-(x^2-4)\le0

Do not omit the modulus of x^2-4. Your equation has to be: |x^2-5x+4|-|x^2-4|\le0
The other method (squaring both the numerator and the denominator) is OK.

ehild

 

[Sumedh]

Thank you very much i got the answer:)
is it easy to put random values before, between and after the zeros to check the sign
or to make the sign table(attached)??

 

[HallsofIvy]

Homework Statement

Solve \frac{|x^2-5x+4|}{|x^2-4|}\le1

Homework Equations

The Attempt at a Solution

as

|x^2-4|will be positive always

cross multiply and take 1 to other side of equation
solve by taking LCM
we get
|x^2-5x+4|-(x^2-4)\le0

How did |x^2- 4| suddenly become x^2- 4?

on solving we get

(x^2-5x+4)-(x^2-4)\le0 and -(x^2-5x+4)-(x^2-4)\le0

the other method I know is to square to remove the modulus function
(x^2-5x+4)^2-(x^2-4)^2\le0


among these which method is correct?
the second method becomes equation of degree 4 i.e.x^4...


please provide hints.

 

[Sumedh]

as it is in modulus it will be positive for any real value of x

if i am wrong please explain me?