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Expanding and factorising

I have a question

100(p+1/2)^2-4(q+1/2)^2

How do I solve this question.

I know that I should square the bracket.

But i'm not too sure about the 1/2 squared. I think it becomes 1/4

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Reply 1
This is not a question, it is an expression. What are you trying to do with it?
Reply 2
Original post by Bobifier
This is not a question, it is an expression. What are you trying to do with it?



factorise
Original post by zed963
factorise


a(b+c)+d(b+c)=(a+d)(b+c)

Does this help?

EDIT: No it doesn't, because I've been a total plonker and misread the q as another p.
(edited 11 years ago)
Reply 4
Original post by Contrad!ction.
a(b+c)+d(b+c)=(a+d)(b+c)

Does this help?


the value in the bracket is different not the same for both
Original post by Contrad!ction.
a(b+c)+d(b+c)=(a+d)(b+c)

Does this help?

EDIT: No it doesn't, because I've been a total plonker and misread the q as another p.


BTW the vid in your sig only looks good cuz she tlaks fast.

She is wrong bout the very first thing with .2222

She adds to the left side, yet she doesn't do the same on the RHS.

It should be X+2 not 2X; she has multiplied by 2 on one side and added 2 on the other.

That is wy her proof that the proof is wrong, is in fact wrong!
Original post by GreenLantern1
BTW the vid in your sig only looks good cuz she tlaks fast.

She is wrong bout the very first thing with .2222

She adds to the left side, yet she doesn't do the same on the RHS.

It should be X+2 not 2X; she has multiplied by 2 on one side and added 2 on the other.

That is wy her proof that the proof is wrong, is in fact wrong!


And at the end she says 'April Fools', it was a piss take.

Original post by zed963
the value in the bracket is different not the same for both


Oh yeah, sorry. I sort of skimread, the p and q looked the same. Argh.

Back to the question - yeah, (12)2=14\left( \frac{1}{2} \right) ^2 = \frac{1}{4}

Do you know how to multiply brackets?
Reply 7
difference of two squares?
Reply 8
Original post by zed963
I have a question

100(p+1/2)^2-4(q+1/2)^2

How do I solve this question.

I know that I should square the bracket.

But i'm not too sure about the 1/2 squared. I think it becomes 1/4


Why do you not learn from the advice you have been given?

The question should be "Simplify the expression" or "Factorise the expression"

I posted a fairly similar answer in your other thread last night.
Reply 9
Original post by steve2005
Why do you not learn from the advice you have been given?

The question should be "Simplify the expression" or "Factorise the expression"

I posted a fairly similar answer in your other thread last night.



My bad
Reply 10
Original post by zed963
I have a question

100(p+1/2)^2-4(q+1/2)^2

How do I solve this question.

I know that I should square the bracket.

But i'm not too sure about the 1/2 squared. I think it becomes 1/4

yes,1/22=1/4because11=1and22=4yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

So, here's the answer:
100(p2+1/4)4(q2+1/4)[br]=100p2+254q21[br]=100p24q2+24100(p^2+1/4)-4(q^2+1/4)[br]=100p^2+25-4q^2-1[br]=100p^2-4q^2+24
Reply 11
Original post by zed963
x

i'm guessing this is gcse maths?
Reply 13
Original post by nm786
yes,1/22=1/4because11=1and22=4yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

So, here's the answer:
100(p2+1/4)4(q2+1/4)[br]=100p2+254q21[br]=100p24q2+24100(p^2+1/4)-4(q^2+1/4)[br]=100p^2+25-4q^2-1[br]=100p^2-4q^2+24


So in order to factorise this it becomes 4(25p2q2+6)[br]=4(5p+q+2)(5pq+3q)[br][br]4(25p^2-q^2+6)[br]=4(5p+q+2)(5p-q+3q)[br][br]
Original post by nm786
yes,1/22=1/4because11=1and22=4yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

So, here's the answer:
100(p2+1/4)4(q2+1/4)[br]=100p2+254q21[br]=100p24q2+24100(p^2+1/4)-4(q^2+1/4)[br]=100p^2+25-4q^2-1[br]=100p^2-4q^2+24


It might be the required answer or it might not. What was the question?

Was the question " simplify " or was the question " factorise ", if the latter, then you answer needs more work.

Edit The following is wrong because it does not follow the previous working.

100(p2+1/4)4(q2+1/4)[br]=100p2+254q21[br]=100p24q2+24100(p^2+1/4)-4(q^2+1/4)[br]=100p^2+25-4q^2-1[br]=100p^2-4q^2+24
(edited 11 years ago)
Original post by nm786
yes,1/22=1/4because11=1and22=4yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

So, here's the answer:
100(p2+1/4)4(q2+1/4)[br]=100p2+254q21[br]=100p24q2+24100(p^2+1/4)-4(q^2+1/4)[br]=100p^2+25-4q^2-1[br]=100p^2-4q^2+24



This contains serious errors

(p + 1/2 )^2 does not equal what you have.
Original post by zed963
I have a question



100(p+12)24(q+12)2100(p+\frac{1}{2})^2 - 4(q+\frac{1}{2})^2

This is just like the one I showed you how to do yesterday

You wil not expand the brackets

It is difference of 2 squares
Original post by zed963
I have a question

100(p+1/2)^2-4(q+1/2)^2

How do I solve this question.

I know that I should square the bracket.

But i'm not too sure about the 1/2 squared. I think it becomes 1/4


There are several methods of factorising the expression.




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Reply 18
Original post by steve2005
This contains serious errors

(p + 1/2 )^2 does not equal what you have.

so what does it equal to?
Original post by nm786
so what does it equal to?


Well

(p+12)2(p+\frac{1}{2})^2


Is certainly not

p2+14p^2 + \frac{1}{4}


So it is up to the OP to see where he has gone wrong

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