What Steps Solve Modulus Inequalities in Algebra?

[Sumedh]

Homework Statement

Solve [tex]\frac{|x^2-5x+4|}{|x^2-4|}\le1[/tex]

Homework Equations

The Attempt at a Solution

as

[tex]|x^2-4|[/tex]will be positive always

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

[tex](x^2-5x+4)-(x^2-4)\le0[/tex] and [tex]-(x^2-5x+4)-(x^2-4)\le0[/tex]

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


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


please provide hints.

 

[rock.freak667]

The Attempt at a Solution

as

[tex]|x^2-4|[/tex]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]

[tex]|x^2-5x+4|-(x^2-4)\le0[/tex]

Do not omit the modulus of x^2-4. Your equation has to be: [tex]|x^2-5x+4|-|x^2-4|\le0[/tex]
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 [tex]\frac{|x^2-5x+4|}{|x^2-4|}\le1[/tex]

Homework Equations

The Attempt at a Solution

as

[tex]|x^2-4|[/tex]will be positive always

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

How did [itex]|x^2- 4|[/itex] suddenly become [itex]x^2- 4[/itex]?

on solving we get

[tex](x^2-5x+4)-(x^2-4)\le0[/tex] and [tex]-(x^2-5x+4)-(x^2-4)\le0[/tex]

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


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


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?