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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Hi,

Can somebody show me a ressource of how to extract the square root of a polynomial that is a square and the same goes for a polynomial that is a cube (to find its cube root)

Thank you!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Can we see the problem?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

I don't have a problem exactly. Unless you want an example because you don't understand what I wrote ???

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

An example would help.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Ok, well if I gave you : (a+b-c)(a+b-c) or (a+b-c)(a+b-c)(a+b-c), each in developped form and I asked you what were the square root of the first and the cube root of the second, you would have : (a+b-c) and (a+b-c).

I want to extract the root in developped form!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You mean like this:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Yes. I think that to find the square root we should use the square root algorithm for numbers.... But I'm not sure.The same goes for the cube root, maybe we should use the cube root algorithm.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Do you have access to some computing power? Do you program or have a CAS?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Nah, I don' even know what is a CAS. But I would want to do it by hand....

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

It will be tedious.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Yes, I know, but I would still like to see how it's done ^^ By curiosity

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

One way would be by a template:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Ok, byt is there a sort of long division method we could use ???

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I would guess so but you would have to know the form of the cube root before you start.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Ah, i was trying to find a way to find the root with only the developped form.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I would use the template provided in post 12. It is a general solver for all trinomials.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Well, I think I could show you an example of what I was talking about. I read it somewhere :

We have :

a^2+2ab+b^2

Now we know that the first term of the root will be "a"

so...

a^2+2ab+b^2(a

"a" squared cancels the other a^2

2ab+b^2(a

Now, we know that 2ab+b^2=(2a+b)b and so if we can figure out a divisor, we will also know the root. We also know the first term of the divisor because it is always the double of the firm term of the root. (Look at the square root algorithm :http://www.basic-mathematics.com/square-root-algorithm.html)

2ab+b^2(a+b (we find, by luck I guess, that we have "b" for the second term.)

____________

2a+b)2ab+b^2

2ab+b^2

______________

0

Anyway, i was searching for something like what I described.

*Last edited by Al-Allo (2014-07-11 03:41:15)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Can you extend that to a cubed form?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

bobbym wrote:

Can you extend that to a cubed form?

Well, I don't know. Let's hope that we can....

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

It is easier to use this:

1) Look at the coefficient in front of a^3 in my example. It is 8. Take the cube root of it. So n1 = 2. Follow so far?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

bobbym wrote:

It is easier to use this:

1) Look at the coefficient in front of a^3 in my example. It is 8. Take the cube root of it. So n1 = 2. Follow so far?

yes

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

2) Look at the coefficient in front of c^3 in my example. It is -27. Take the cube root of it. So n3 = -3.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

bobbym wrote:

2) Look at the coefficient in front of c^3 in my example. It is -27. Take the cube root of it. So n3 = -3.

Following

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

3) Now take a look at the abc term. What do you see?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

bobbym wrote:

3) Now take a look at the abc term. What do you see?

That we have 36 ???

*Last edited by Al-Allo (2014-07-11 03:58:48)*

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