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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

Hello sir, please assist me to solve these algebra.

x+x+x2-x2. please the two **x** are squared, I cannot locate them(the squares) in this forum in order to apply them conveniently on them.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi EbenezerSon;

Welcome to the forum.

This looks like what you want done:

x + x = 2x

2x + x^2 = 2x + x^2

Now x^2 - x^2 = 0.

2x + 0 = 2x

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

Thank you sir, God bless, but please one more problem from algebra.

1. (h^2 - h^2) -p(h + h).

2. X^2 + X^2 + X^2 - Y^2

Please, where can I use the mathematical items, like the (square), (plus symbol) and the others in this forum, as you have used them nicely above.

Thanks.

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

You can use latex, that is what I am using. It is all done by a site. Take a look here.

http://latex.codecogs.com/editor.php

1) (h^2 - h^2) -p(h + h)

h^2 - h^2 is 0

0 - p(h + h)

Can you do the next step?

2) X^2 + X^2 + X^2 - Y^2

X^2 + X^2 + X^2 is 3X^2

3X^2 - Y^2 we can not simplfy any further so you are done.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

0-p(h + h)

= -hp - hp.

I know only one thing - that is that I know nothing

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

But sir, the book in which I am learning this from has given me a different answer altogether.

It is saying, though an algebra but this type of question is called "difference of two squares".

I would provide his procedure here, which I could not quite comprehend some aspects of his procedure.

I really appreciate your enormous assistance.

Thanks.

*Last edited by EbenezerSon (2013-07-05 10:50:17)*

I know only one thing - that is that I know nothing

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,531

Hi, if you want to simplify 0-p (h+h), it would be ph + ph which is 2ph. Secondly, difference of two squares implies that you need to factor these, not just simplify. To do that, you must use the identity (a^2)-(a^2) equals (a+b)(a-b).

*Last edited by Shivamcoder3013 (2013-07-05 14:09:40)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

But how can you factor them?

1) (h^2 - h^2) -p(h + h)

h^2 - h^2 = 0

There is no difference of 2 squares.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Ebenezerson I think u are posting wrong question

There are 10 kinds of people in the world,people who understand binary and people who don't.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

Regarding the problem above I think the member omitted the 'negative sign'.

The following is how I see it;

0-p(h+h) = (-ph - ph)

= -2ph , What do you say with my solution? The 'negative sign' was affecting the P.

I know only one thing - that is that I know nothing

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

(h^2 - h^2) -p(h+h)

= (h - h)(h + h) -p(h+h)

= (h + h)(h - h - p) . The final solution provided by the book.

*Last edited by EbenezerSon (2013-07-08 02:23:57)*

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

That is an extremely weird answer they give.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 456

bobbym wrote:

Hi;

That is an extremely weird answer they give.

Sir, forgive me I should have provided how the question should be solved. the question says:

(The factorize the expression completely). I think it should be the answer if we are to factorize it, I have now comprehended it. And I can explain.

What do you say.

*Last edited by EbenezerSon (2013-07-08 02:35:24)*

I know only one thing - that is that I know nothing

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

If you now understand what they did then that is okay with me.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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