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

Last edited by rickyoswaldiow (2005-12-06 03:04:45)

.93968

This took four iterations using Newton's Method.

x - f(x)/f'(x)

I learnt it as the Newton-Raphson process, but whatever.

Basically, it is a way to estimate the root of an equation that is too complicated to solve easily.

f(x) represents the function of x, in this case being 4x³ - 3x - 1/2, and f'(x) is the differential of the function, in this case being 12x² - 3.

So, the equation is x - (4x³ - 3x - 1/2) ÷ (12x² - 3)

You use this by picking a starting value for x and substituting that into the above equation. The value it returns should be closer to the root than the value that you used. You can then use the new value to get an even better estimation until you have an answer to the required degree of accuracy.

So, if we start with x_0 = 1:

x_1 = 1 - (4 - 3 - 1/2) ÷ (12 - 3) = 0.9444...
Now we use this value and do it again.

x_2 = 0.9444... (4*(0.9444...)³ - 3*0.9444... - 12) ÷ (12*(0.9444...)² - 3)
      = 0.9397...

x_3 = 0.9397...

Now that we have 2 values that are the same to 4 decimal places, we can stop and call that the answer. irspow used another iteration, but that was because he decided to be nice and give you an extra decimal place, free of charge.

I think that

Last edited by krassi_holmz (2005-12-11 09:24:49)