Absolute Values (1 Viewer)

[YBK]

hey..
i'm having troubles solving absolute values such as |x-4| > 2x

is this the correct way to do it?

Restriction x >= 0

(x-4)^2 > (2x)^2
0 > (2x)^2 - (x-4)^2
0 > (3x-4)(x+4)

x= 3/4 or x = -4
But x = -4 does not fit the restriction

Therefore 0 <= x < 4/3


Ummm.. am i right... i'm not sure of the x>=4 thing...

Thanks!

 

[Trebla]

Restriction x >= 0

(x-4)^2 > (2x)^2
0 > (2x)^2 - (x-4)^2
0 > (3x-4)(x+4)

x= 3/4 or x = -4

(x-4) < -2x
= -4 < -3x
= x > 3/4

OMG, you both got it wrong! Check your algebra, the number should by 4/3, NOT 3/4. YBK, the question asks for an inequality, I don't know why you used an equality instead. Also, kadlil, you forgot to reverse your inequality signs because you divided by a negative number.

I don't think there should be a restriction, because you are dealing with inequalities and not equalities.
Sketch y = |x - 4| and y = 2x and find the domain where the absolute value graph is ABOVE the straight line. You'll have a ROUGH idea on where your solutions would lie.
Algebraically, as kadlil demonstrated should be the correct method and without the errors, you should end up with:
x < - 4 AND x < 4/3

The COMBINED domain of solutions turns out to be x < 4/3, which should be your final answer.

 

[Trebla]

To help you in future questions, you should remember this rule:
For the equation:
|a| > b
the solutions are:
a > b
AND
a < - b

For the equation:
|a| < b
the solutions are:
- b < a < b