#1
I posted in the M/S help thread, but I didnt get a response yet. I know the pit isnt my teacher, and I would wait, but Im going to sleep in some time and I have a test tomorrow.
Help please?



Can I safely square in inequalities on my test tomorrow?

Or are there certain rules? (Which im sure there are )

Edit: And Im confused about a concept, so can someone solve this sum for me? (made it up right now)

|x|+3>|x^2-x|

Without using squaring method or calculus?


Greatly appreciated if a mod leaves this open as a favour to me. [Why should you give a shit about me? Well you shouldnt, but please ]
#3
If you could graph it that would make it a whole lot easier. Algebraically that's just annoying to solve. Basically each absolute value term has two possible values 1) itself 2) its negative. So |x^2-x| could be x^2-x or -x^2+x and you have to go through each combination thereof in the solution.
#4
I don't know what the question is to be honest? But that little equation at the bottom is only working with 0, 1 and 2.
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#6
Hmm, I'm really not sure how to go about this one :/ try taking them all over to one side?
Blog Of Awesome UGers.
Quote by OddOneOut
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#7
Quote by Venice King
Hmm, I'm really not sure how to go about this one :/ try taking them all over to one side?


This involves absolute values, and Im squaring to get rid of them, so that wouldnt work
#8
Quote by luv090909
^

a>b

can i square them and make it
a^2 > b^2

Well no, you can't.
Say a = 3 and b= -4, then a>b.
But it wouldn't be right to say a^2 > b^2 ( as 9 < 16)
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Homer: Fun too!

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#9
Quote by circus musician
Well no, you can't.
Say a = 3 and b= -4, then a>b.
But it wouldn't be right to say a^2 > b^2 ( as 9 < 16)



I was looking for rules on squaring.

Got any ideas?

Google doesnt help much
#10
Quote by luv090909
I was looking for rules on squaring.

Got any ideas?

Google doesnt help much

Based on my example, I would have to say I don't believe that squaring would help at all. From what I can remember, I think you have to use graphs to do this but it's been a while since I did it.
Marge: You know Homer, it's easy to criticize.
Homer: Fun too!

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#11
|x|+3>|x^2-x|
3>|x^2-x|-|x|
3>|x|*|x-1|-|1|
4>|x|*|x-1|
4>|x(x-1)|
-4<x(x-1)<4
0<x<2.56

The rules used are: |a|/|b|=|a/b|
and |a|<b is the same as -b<a<b if b positive