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Expanding and factorising

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

    factorise
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    (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.
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    (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
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    (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!
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    (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, \left( \frac{1}{2} \right) ^2 = \frac{1}{4}

    Do you know how to multiply brackets?
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    difference of two squares?
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    (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.
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    (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
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    (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, 1/2^2=1/4 because 1*1=1 and 2*2=4

    So, here's the answer:
    100(p^2+1/4)-4(q^2+1/4)

=100p^2+25-4q^2-1

=100p^2-4q^2+24
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    (Original post by zed963)
    x
    i'm guessing this is gcse maths?
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    (Original post by zed963)
    My bad

    http://www.thestudentroom.co.uk/show...=#post37257318
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    (Original post by nm786)
    yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

    So, here's the answer:
    100(p^2+1/4)-4(q^2+1/4)

=100p^2+25-4q^2-1

=100p^2-4q^2+24
    So in order to factorise this it becomes 4(25p^2-q^2+6)

=4(5p+q+2)(5p-q+3q)
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    (Original post by nm786)
    yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

    So, here's the answer:
    100(p^2+1/4)-4(q^2+1/4)

=100p^2+25-4q^2-1

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

=100p^2+25-4q^2-1

=100p^2-4q^2+24
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    (Original post by nm786)
    yes, 1/2^2=1/4 because 1*1=1 and 2*2=4

    So, here's the answer:
    100(p^2+1/4)-4(q^2+1/4)

=100p^2+25-4q^2-1

=100p^2-4q^2+24

    This contains serious errors

    (p + 1/2 )^2 does not equal what you have.
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    (Original post by zed963)
    I have a question
    100(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
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    (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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    (Original post by steve2005)
    This contains serious errors

    (p + 1/2 )^2 does not equal what you have.
    so what does it equal to?
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    (Original post by nm786)
    so what does it equal to?
    Well

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


    Is certainly not

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


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

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Updated: April 23, 2012
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