Absolute values (1 Viewer)

[Aznmichael92]

Hey I have trouble solving this question. Can anyone help please? Thanks

l 4x - 1 l > 2 x rt[x(1-x)]

 

[tommykins]

square both sides.

 

[Aznmichael92]

thank you very much I forgot about that method

 

[Aznmichael92]

um i think i did an error or something but not too sure. Can someone check if I did it correctly

l 4x - 1 l > 2 x rt[x(1-x)]
Square both sides

16x^2 - 8x + 1 > 4x - 4x2

 

[tommykins]

i'm confused whether the 2 x rt[x(1-x)] is 2 multiplied by rt etc. or 2x.rt etc.?

 

[Trebla]

um i think i did an error or something but not too sure. Can someone check if I did it correctly

l 4x - 1 l > 2 x rt[x(1-x)]
Square both sides

16x^2 - 8x + 1 > 4x - 4x^2

Looks correct. Now just solve for x. You should get x > 1/2 and x < 1/10.
BUT there is a trick to this question!
The number inside the square root MUST BE NON-NEGATIVE!!!
i.e. x(1-x) > 0
=> 0 ≤ x ≤ 1
This means that the actual solution is 0 ≤ x < 1/10 and 1/2 < x ≤ 1