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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

bob bundy wrote:

At first, I did not spot that this is the difference of two squares. But then I noticed that (q^2 - 6qr + 9r^2) is a perfect square:

So try this:

put P = 3p and Q = (q - 3r) and use

Bob

Last but not the least could you help me with an illustration, how you got the (q - 3r)^2 ?. :-)

*Last edited by EbenezerSon (2013-07-10 09:11:04)*

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**bob bundy****Moderator**- Registered: 2010-06-20
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OK. Here's an example of a perfect square. I work best with pictures so I've made a picture below for this example.

Could this be a perfect square ?

Well it has x^2. That's a good start. And it has 9y^2 = (3y)^2 so it's looking like it might be a perfect square.

Have a look at my picture. Click it to make it bigger.

With a perfect square there are four boxes.

Two are squares: x^2 and (3y)^2

The other two boxes have to be the same rectangle shape, once one way round, and then again turned 90 degrees.

So can the third term in the expression, 6xy, be made into two equal rectangles ... it must be 3xy and 3xy.

So now I can factorise the expression into two equal brackets:

For your example, you had

q^2 tick

9r^2 = (3r)^2 tick

Split the middle term in two: -6qr = -3qr - 3qr tick

so

Now I shall have to log out so here's a hint for number (4)

Three of these terms will make a perfect square. You need to decide which three.

Then choose P and Q and use the difference of two squares.

I'll check tomorrow to see how you are getting on.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

4a^2 - 12ab - c^2 + 9b^2

= 4a^2 - 12ab - 9b^2 - c^2

= 4a^2 - 6ab - 6ab + 9b^2

= 2a(2a - 3b) -3b(2a - 3b)

= (2a - 3b)(2a - 3b) = (2a - 3b)^2.

------------------------------------------

-----------------------------------------

(2a - 3b)^2 - c^2 = (2a - 3b - c)(2a - 3b + c). Is the final answer correct?

*Last edited by EbenezerSon (2013-07-11 04:21:20)*

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

The book gave this as the final answer:

(2a - 3b - c)(2a - 3b '- C')

I don't understand how it got the '-C', I think it should be '+C' instead.

What do you say?

*Last edited by EbenezerSon (2013-07-11 00:05:36)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

Your answer in post 28 is correct. don't know what the book is on about.

Got to go out. I'll check again later.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

hi EbenezerSon

Just got back home. There's a way to check algebra when you want to know if you've done it correctly. Give me a moment and I'll post what you have to do.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

bob bundy wrote:

hi Ebenezerson

I think you'll need to multiply these brackets out like this

Then you can start to simplify.

Can you fill in the dots ?

Alternative way to do this:

difference of two squares:Have you met this before ?

If so, you could put p = 2m + 2k and q = m - 2k

Bob

Sir this one, I tried resolving it my self with my understanding, but got different result, please I would be much obliged if you would take the pain to explain it to my understanding once more by manipulating each step. Please.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

OK. here we go.

The rules of algebra are the same as the rules of arithmetic. So you can check your algebra by substituting some numbers.

Let's use Q4 as an example.

You started with 4a^2 -12ab - c^2 + 9b^2

You've got one answer. The book has another. Which is right?

Choose numbers for a, b, and c. It's best to choose different numbers and not 0 or 1. So I'll make

a = 2

b = 3

c = 5

All different with no common factors.

The original expression is 4a^2 -12ab - c^2 + 9b^2 = 4 x 2 x 2 - 12 x 2 x 3 - 5 x 5 + 9 x 3 x 3 = 16 - 72 - 25 + 81 = 97 - 97 = 0

Wow! I didn't plan that to be zero. It was just a lucky chance!

Now your final answer.

(2a - 3b - c)(2a - 3b + c) = (2 x 2 - 3 x 3 - 5) x ( 2 x 2 - 3 x 3 + 5) = (4 - 9 - 5) x (4 - 9 + 5) = -10 x 0 = 0

This is the same as what you started with.

Book answer:

(2a - 3b - c)(2a - 3b - C) = (2 x 2 - 3 x 3 - 5) x (2 x 2 - 3 x 3 - 5) = (4 - 9 - 5) x (4 - 9 - 5) = -10 x -10 = 100.

This isn't correct. The book answer must be wrong.

There's a small chance that you made two errors that 'cancelled out' but it's unlikely.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

Sir this one, I tried resolving it my self with my understanding, but got different result, please I would be much obliged if you would take the pain to explain it to my understanding once more by manipulating each step. Please.

what was your answer?

Bob

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

Errr sir, that's good, so always to check, should I pick the numbers randomly in my head to do the substitution? To see if will get (0) as result for both? Or the figure I will get should be the same in each case? And not necessarily (0)?

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

I got:

3m^2 - 8km + 13k^2, Which is impossible for me to split the middle term.

Please help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,358

Hi EbenezerSon;

Do you need to factor that?

Because there is always a factorization of an expression like that with a lot of work and help from a computer to check the expansions this is what I found:

This involves complex numbers and I am sure this is not what they want so have you copied the problem correctly?

Here are some easy to factor possibilities.

**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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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

Ebenezerson wrote:

I got:

3m^2 - 8km + 13k^2, Which is impossible for me to split the middle term.

I showed you this one back in post 16.

The -8km term should have been +12mk + 4mk so it looks like you've made two sign mistakes.

The 13k^2 should have been 9k^2 - 4k^2 so once again a sign mistake.

I recommend you do some practice on expanding brackets with some negative signs. You find this page and exercises helpful:

http://www.mathsisfun.com/algebra/expanding.html

Errr sir, that's good, so always to check, should I pick the numbers randomly in my head to do the substitution? To see if will get (0) as result for both? Or the figure I will get should be the same in each case? And not necessarily (0)?

Yes, you can pick random numbers. You may miss a mistake if you choose the same number for 'a' and 'b' so pick different numbers. It was a complete chance that the expression came out as 0. With a different choice for a, b and c it would have produced a completely different result. What you are looking for is the **same result** from both expressions.

Let's use your other question as an example.

(1) (2m + 3k)^2 - (m - 2k)^2.

Let's choose m = 2 and k = 3

(2m + 3k)^2 - (m - 2k)^2 = (2 x 2 + 3 x 3)^2 - (2 - 2 x 3)^2 = (4 + 9)^2 - (2 - 6)^2 = 13^2 - (-4)^2 = 169 - 16 = **153**

So the correct answer should come to **153**.

Your answer 3m^2 - 8km + 13k^2 = 3 x 2 x 2 - 8 x 2 x 3 + 13 x 3 x 3 = 12 - 48 + 117 = 129 - 48 = 81 so this must have an error.

My result 3m^2 + 5k^2 + 16mk = 3 x 2 x 2 + 5 x 3 x 3 + 16 x 2 x 3 = 12 + 45 + 96 = **153**

Bob

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

I have understood #38 very well, but please, see how I came by that final answer, I have tried to understand how you solved it at #19 but still I haven't grasp it.

*Last edited by EbenezerSon (2013-07-12 07:57:46)*

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

(2m^2 + 3k)^2 - (m - 2k)^2 . Please see how I solved it, and correct where I have made the mistakes so that I could identify it. Please.

Here I go:

(2m + 3k)^2 - (m - 2k)^2 = (2m + 3k)(2m + 3k) - (m - 2k)(m - 2k) . by expansion.

= Multiplying the first expansion > (2m + 3k)(2m + 3k) = 4m^2 + 6mk + 6mk + 9k^2.

Multiplying the second expansion > (m - 2k)(m - 2k) = m^2 - 2mk - 2mk + 4k^2.

= (4m^2 + 6mk + 6mk + 9k^2) - (m^2 - 2mk - 2mk + 4k^2)

= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2

= adding them > 3mk^2 + 12mk - 4mk + 13k^2

= 3mk^2 + 8mk + 13k^2.

Please this is how I got my answer, I know I am wrong but I cant help.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

(2m + 3k)^2 - (m - 2k)^2 = (2m + 3k)(2m + 3k) - (m - 2k)(m - 2k) . by expansion. Correct!

= Multiplying the first expansion > (2m + 3k)(2m + 3k) = 4m^2 + 6mk + 6mk + 9k^2. Correct!

Multiplying the second expansion > (m - 2k)(m - 2k) = m^2 - 2mk - 2mk + 4k^2. Correct!

= (4m^2 + 6mk + 6mk + 9k^2) - (m^2 - 2mk - 2mk + 4k^2) Correct!

= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2 This is where the mistake is.

**-(m^2 - 2mk -2mk + 4k^2) = -m^2 + 2mk + 2mk - 4k^2**

This is -1 times the bracket, so **every term changes sign.**

= adding them > 3mk^2 + 12mk - 4mk + 13k^2 Should be 3m^2 + + 12mk **+ 4mk **+ 9k^2 **- 4k^2**

= 3mk^2 + 8mk + 13k^2. Should be 3m^2 + 16mk - 5k^2

I recommend that you practise expanding brackets with negatives.

http://www.mathsisfun.com/algebra/expanding.html

This will help you with problems like this one.

Bob

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

bob bundy wrote:

= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2 This is where the mistake is.

-(m^2 - 2mk -2mk + 4k^2) = -m^2 + 2mk + 2mk - 4k^2[/b]

This is -1 times the bracket, so

every term changes sign.

Thank you very much indeed Sir!

Errrr! That was where the mistakes stem from, candidly speaking I had thought, I one has to take off the bracket, after multiplication has taken place within the bracket.

I have now grown wary to multiply any sign outside the bracket after multiplcation has taken place within the bracket, especially negative signs.

Thanks Sir for the provision of the link, God bless.

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 438

Simplify the following:

Please help me solve this. (x^3/2 + x^1/2) (x^1/2 - x^-1/2)/(x^3/2 - x^1/2)^2. I got x^2 - x + x/x^3 - x^2 - x^2 as the final answer. The book has x + 1/x(x - 1) as its answer which I don't comprehend. The book did not show the procedures.

Thanks in advance.

*Last edited by EbenezerSon (2013-07-18 02:47:33)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,358

Hi;

Is this the way the problem looks?

**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: 438

This is how the top right bracket looks: (x^1/2 - x^-1/2) The last X raises to the power negative one over two(x^-1/2).

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,358

Hi;

Those are both the same. But to put it into the form you want then we have?

**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: 438

Errr okay, I see!

But Sirs please solve for me.

Thanks in advance.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,358

Hi;

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: 438

Please I want to follow the procedure before getting the answer, the book gave me exactly your answer, but the workings was not shown.

Thanks once more.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,358

I multiplied the numerator out and then the denominator. Just plain multiplication.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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