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    Hey there, I've been looking through my maths work that we've done so far (C1 obviously) and I'm up to completing the square and I just don't get what to do, I can't work out what I'm doing from my own notes. I looked at a textbook and still don't really get it.

    Theres a few different types of questions from the textbook, the questions sort of build up as you go through it (core maths for advanced level)

    So the first set say: Add a number to each expression so that the result contains a perfect square as a factor. (I don't really know what I'm supposed to be doing)
    Heres an e.g. of a question: x^2 - 4x, another is 2x^2 - 4x, and so on.

    But here are by the looks of it the main type of question: Solve the equations by completing the square, giving the solutions are in surd form:
    I'll try and give a variation of questions...

    1. X^2 + 8x = 1
    2. 2X^2 - x - 2 = 0
    3. 2x^2 + 4x = 7

    But yeah, if anyone could show me one of those examples with all the steps explained I'd be grateful, because I really don't know what I'm doing! What is actually meant by completing the square?
    The answers are in the back of the textbook but of course there are no steps so its pretty useless for me to write the answer straight down.
    Thanks a lot
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    1. X^2 + 8x = 1
    X^2 + 8x - 1 = 0 [-1 from both sides]
    [[x + 4]^2 -16] - 1 = 0 [factorising X^2 + 8x and subtracting the remainder created by 4*4]
    [x+4]^2 - 17 = 0 Done [completing the fraction.. ie -16-1 to get the final answer]

    Basically.. look at this part: X^2 + 8x and factorise it, ie [x+4]^2, but then that creates a +16 that you don't want (since you must keep it the same as the original equation]

    For question two... look at 2X^2 - x, and factorise it ie [x-1/2]^2... see if you can do the rest. [you get a +1/4 but you need -2 so you will have to subtract something]
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    1. x^2+8x=1.

    first put the 1 on the end the other side of equation, so x^2+8x-1=0.

    then put (x^2+the coefficiant of x divided by 2), so (x^2-4x)^2.

    then double coefficient of x in brackets and change sign, placing outside bracket,
    so (x^2-4x)^2+8, and then -1=0 on the end (from the top equation).

    then simplify. (x^2-4x)^2+7=0.

    take the coefficiant of x in the bracket and change the sign to get the first value of x, so it is 4.

    then take the number outside the bracket, so 7.

    x= 4 or 7.
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    okay ignore my comment, i forgot a crucial stage!!! whoopsie.
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    Oh and... [x+4]^2 - 17 = 0

    To find x... you just solve in a regular way... [x+4]^2 = 17 etc
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    The Song
    This is how we complete the square
    Complete the square
    Complete the square
    This is how we complete the square
    Of a simple quadratic function

    Half the coefficient of x
    efficient of x
    efficient of x
    Half the coefficient of x
    Then subtract its square from the constant.

    What it actually means
    So, always arrange the equation into the form like 2x^2 + 5x + 13 = 0
    Then divide everything by the coefficient of x^2, to get 2(x^2 + 5/2x + 13/2) = 0
    Then complete the square on the (x^2 + 5/2x + 13/2) part

    "Half the coefficient of x"
    Half of 5/2 is 5/4, right?
    Then put this 5/4 in a bracket like so: (x-5/4)^2. Step one done.

    "Then subtract its square from the constant"
    i.e. subtract (5/4)^2 from the original constant, which was 13/2- thus leaving us with 13/2 - (5/4)^2. Step 2 done.

    Then we add the two parts formed together.
    So x^2 + 5/2x + 13/2 = (x-5/4)^2 + (13/2 - (5/4)^2)

    Since the equation was originally doubled, then
    2x^2 + 5x + 13
    = 2(x^2 + 5/2x + 13/2)
    = 2((x-5/4)^2 + (13/2 - (5/4)^2))
    = 2(x-5/4)^2 + 2(13/2 - (25/16))
    = 2(x-5/4)^2 + 13 - (25/8)
    = 2(x-5/4)^2 + 79/8

    And we've just completed a hard "complete the square" question!
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    (Original post by Icy_Mikki)
    The Song
    This is how we complete the square
    Complete the square
    Complete the square
    This is how we complete the square
    Of a simple quadratic function

    Half the coefficient of x
    efficient of x
    efficient of x
    Half the coefficient of x
    Then subtract its square from the constant.

    What it actually means
    So, always arrange the equation into the form like 2x^2 + 5x + 13 = 0
    Then divide everything by the coefficient of x^2, to get 2(x^2 + 5/2x + 13/2) = 0
    Then complete the square on the (x^2 + 5/2x + 13/2) part

    "Half the coefficient of x"
    Half of 5/2 is 5/4, right?
    Then put this 5/4 in a bracket like so: (x-5/4)^2. Step one done.

    "Then subtract its square from the constant"
    i.e. subtract (5/4)^2 from the original constant, which was 13/2- thus leaving us with (5/4)^2 - 13/2. Step 2 done.

    So x^2 + 5/2x + 13/2 = (x-5/4)^2 + (13/2 - (5/4)^2)

    Since the equation was originally doubled, then
    2x^2 + 5x + 13
    = 2(x^2 + 5/2x + 13/2)
    = 2((x-5/4)^2 + (13/2 - (5/4)^2))
    = 2(x-5/4)^2 + 2(13/2 - (25/16))
    = 2(x-5/4)^2 + 13 - (25/8)
    = 2(x-5/4)^2 + 79/8

    And we've just completed a had "complete the square" question!
    :eek: how cool i cant believe some people get to learn maths through songs
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    (Original post by scraceus999)
    okay ignore my comment, i forgot a crucial stage!!! whoopsie.
    Lol exactly what I was thinking!
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    Wow, thats actually pretty easy when broken down! Thanks a lot, can anyone find me any links to questions on CTS? Not ones that involve CTS, but just an equation, or perhaps some of you could give me some?
    Thanks
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    complete the square:

    1. x^2 - 4x -12 =0

    and if you think that's too easy then:

    2. 4x^2 + 17x = 6x - 2x^2
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    The Song
    That should be banned.

    Completing the square is the easiest thing ever. Just think about what happens when you expand something, then you have it.

    I think it better to have a understanding of something on a deeper level, you must be doing something wrong.
 
 
 
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