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**austin81****Member**- Registered: 2005-03-21
- Posts: 39

Pls help me solve for x in 4x^3-3x=1/2. leave answers to 4decimal places. Thanks in Advance

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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 (2005-12-05 04:04:45)*

Aloha Nui means Goodbye.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

4x - 3x = CUBE ROOT OF 1/2?

I thought I knew how to do this, but maybe not.

*Last edited by rickyoswaldiow (2005-12-05 04:07:17)*

Aloha Nui means Goodbye.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

.93968

This took four iterations using Newton's Method.

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

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**naturewild****Member**- Registered: 2005-12-04
- Posts: 30

irspow wrote:

.93968

This took four iterations using Newton's Method.

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

I cannot help but wondering what this formula represent. Care to elaborate?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Nice explanation mathsyperson, I use Newton's method all of the time. It is very useful for difficult problems. I must be showing my age by calling it Newton's Method.

By the way, I didn't compute an extra digit for nothing, I wanted to be sure that the fourth digit did not change if he was going to keep it as accurate.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I think that

*Last edited by krassi_holmz (2005-12-10 10:24:49)*

IPBLE: Increasing Performance By Lowering Expectations.

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**nikcos****Member**- Registered: 2014-10-10
- Posts: 1

I try with this method : Resolving cubic equations by an homography

But I don't understand,

I'm not sure if it works

*Last edited by nikcos (2014-10-10 00:57:01)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,702

Hi;

Which method, looks like there are two? Can you please show what problem you were working on and where you got stuck?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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