Solve: |2x-1|-|x| smaller than or equal to 0 (1 Viewer)

[enigma_1]

|2x-1|-|x| smaller than or equal to 0


How do you know where to put the negative sign and stuff? thanks

 

[Kurosaki]

|2x-1|-|x| smaller than or equal to 0


How do you know where to put the negative sign and stuff? thanks

Consider cases: when x> 1/2, when x<1/2, when x<0, when x>0 and how this affects the behaviour of the absolute value functions.
You could also draw a graph.

 

[Squar3root]

alternatively you could solve |2x-1|= |x| to find the points of intersection and then graph |2x-1|= |x| and see when |2x-1| - |x| is less than or equal to 0

 

[xGhanem]

(2x-1)^2 < (x)^2

4x^2 - 4x + 1 < x^2

3x^2 - 4x + 1 < 0

(3x-1)(x-1) < 0

x is greater or equal to 1/3, but less or equal to 1.

This method (squaring both sides) is longer but I find it easier to understand.

 

[Trebla]

Bring |x| to the other side of the inequality, square both sides (which preserves the sign as both sides are non-negative) and solve the quadratic

 

[braintic]

Bring |x| to the other side of the inequality, square both sides (which preserves the sign as both sides are non-negative) and solve the quadratic

Another solution:

2|x - 1/2| < |x|

|x| / |x - 1/2| > 2

First treat the case of equality, ie. |x| / |x - 1/2| = 2.

Treating the absolute values as distances:
The ratio of the distance of x from zero to the distance of x from 1/2 is 2.
That is a round about way of saying: 'Divide the interval from x=0 to x=1/2 in the ratio 2:1.
You get two answers because the division can be done internally and externally.
This divides the number line into 3 regions.

Then test a point in each region to determine the solution to the inequality.

 

[Ikki]

LOL question done to death haha

 

[braintic]

LOL question done to death haha

Yeah right, LOL, ha .............................. ha ......................... ha ................................... ugh