Math Is Fun Forum

yep, it's a joke.

Sounds like a job for Crazy Holmes

I'd start with x/x^2, firstly I'd "reciprocal" which basically means you take the value with the power (x^2) and move it to the top, when you do this you MUST change the sign of the power:

x / x^2  = x * x^-2

I'd then do the same with the next part of the sum, giving you b - ax * x^-3.  Now all that is left is

(x * x^-2) + (b - ax * x^-3), since none of the values share similar powers, I think this is as far as you can go.  (Don't take my word for it, I'm new to this too).

maybe you'll have heard of the expression "pwned" which is derived from the word "owned".  I recently read a question asking how this word was to be pronounced, many people had many thories.  My theory is that since it is internet slag, the word was not meant to be pronounced (only typed) and thus the internet is spawning a global language that has no pronounciation.  Do any other languages such as this exist?

Thanks again, one final question, the only one left that I am unsure of my workings:

y = sqrt(4/x^3) = sqrt(4x^-3) = 2x^-3/2
dy/dx = -3x^-5/2
Is this right?

in a question like y = 3x^4,  do you multiply the 4 by the 3 to make dy/dx = 12x^3?
i.e.  y=ax^b   dy/dx = (ba)x^b-1

The picture of yours that I see has the hat on! Are you trying to change it back to your normal one?

so you're saying I should divide the whole equation through by the co-efficient of x^2 and then factorise normally, multiplying the co-efficient back in at the end?

Learning maths for the sake of learning maths is probably very dull.  Most people learn maths because they need to for somthing else i.e. software development.
   I've found studying maths can be a little tedious, but I only fail at learning it if I try to understand the maths from the start.  When I am being taught a new method in mathematics, instead of trying to UNDERSTAND what is going on, I just accept the steps I need to take and take them.  The understanding of what's happening usually sinks in when you have been working with somthing for a while.

To solve the equation you just have to remember that whatever you do to one side you must do to the other.  The idea is to get the x on it's own.

so 4x^3 - 3x = 1/2 becomes
    4x^3 = 3x + 1/2

You see that I made the "-3x" on the left equal 0 by adding 3x to it.  Because I added 3x on one side I must add 3x onto the other side too.
   For the next step you must take that 4 off from in front of the x^3  (that 4 means multiply so you must divide 4x^3 by 4 to get it as x^3).  Again, since you have divided 4 on one side, so you must on the other.

x^3 = (3x + 1/2)  / 4

   Finally, the opposite of "cubing" a number (which is what the ^3 does) is finding it's cube root.  so x = CUBE ROOT OF (3x+1/2)/4
          -2x = CUBE ROOT OF 1/8
   You'll probably need a calculator for that last step