Maths Q (1 Viewer)

[azureus88]

I was doing this question and i got to a point where i found that (1-2k)>0.5. I need to find an inequality with (1-2k)/k^2 in it given 0<k<0.25.

Can i just go from the first step to [(1-2k)/k^2]>8 or do i need some sort of explanation. If so, what should i write?

 

[vds700]

I was doing this question and i got to a point where i found that (1-2k)>0.5. I need to find an inequality with (1-2k)/k^2 in it given 0<k><0.25.

Can i just go from the first step to [(1-2k)/k^2]>8 or do i need some sort of explanation. If so, what should i write?

This doesnt make sense

how exacrly does
1 - 2k > 5 become
(1 - 2k)/k² > 8????

Maybe post the whole question
Btw whats "0<0.25"? meant to be
</k>

EDIT: Just seen Lolokays post, i misread the question

 

[lolokay]

1-2k>1/2
2k<1/2
k<1/4
k²<1/16
8k²<1/2
1-2k>8k²
(1-2k)/k² > 8

assuming 0 < k < 0.25 is what you meant

 

[Trebla]

Show all working, as it is not obvious how 1 - 2k > 1/2 leads to (1-2k)/k² > 8.