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    How do you do it?.
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    (Original post by BTREDinnick)
    How do you do it?.
    Most questions will be in the general form
    ax^2+bx+c
    for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

    e.g. x^2+4x+5

    1. half the b value and re-write like this
    so 4/2 is 2 and this is what you put inside the bracket.

    (x+2)^2 +5

    2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
    like,
    (x+2)^2 -4 + 5

    You can do the first two steps all at once to save yourself time.

    3. Tidy up the 'normal' numbers at the end so -4+5 = 1

    so at the end it will look like this

    (x+2)^2 +1


    For when there is a coefficient of x^2 which is larger than 1, you do this
    Question : 2x^2 + 8x +9
    1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
    2(x^2 + 4x) + 9

    2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
    You will need 2 sets of brackets
    2((x^2 + 2) -4) +9

    3. Expand the first bracket
    The minus 8 is from the 2 x -4
    2(x^2 + 2) -8 +9

    4. And tidy tidy,

    2(x^2 + 2) + 1
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    Thanks. That helps me alot.
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    (Original post by zara_ruby)
    Most questions will be in the general form
    ax^2+bx+c
    for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

    e.g. x^2+4x+5

    1. half the b value and re-write like this
    so 4/2 is 2 and this is what you put inside the bracket.

    (x+2)^2 +5

    2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
    like,
    (x+2)^2 -4 + 5

    You can do the first two steps all at once to save yourself time.

    3. Tidy up the 'normal' numbers at the end so -4+5 = 1

    so at the end it will look like this

    (x+2)^2 +1


    For when there is a coefficient of x^2 which is larger than 1, you do this
    Question : 2x^2 + 8x +9
    1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
    2(x^2 + 4x)^2 + 9

    2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
    You will need 2 sets of brackets
    2((x^2 + 2)^2 -4) +9

    3. Expand the first bracket
    The minus 8 is from the 2 x -4
    2(x^2 + 2)^2 -8 +9

    4. And tidy tidy,

    2(x^2 + 2)^2 + 1
    I forgot to put some of the x^2's outside the brackets!
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    (Original post by zara_ruby)
    Most questions will be in the general form
    ax^2+bx+c
    for basic questions in normal maths where there the coefficient of x^2 is 1 you do this;

    e.g. x^2+4x+5

    1. half the b value and re-write like this
    so 4/2 is 2 and this is what you put inside the bracket.

    (x+2)^2 +5

    2. Then you take the 2 inside the bracket, square it and multiply it by -1 and put it outside the bracket. 2^2 = 4 x -1 = =4
    like,
    (x+2)^2 -4 + 5

    You can do the first two steps all at once to save yourself time.

    3. Tidy up the 'normal' numbers at the end so -4+5 = 1

    so at the end it will look like this

    (x+2)^2 +1


    For when there is a coefficient of x^2 which is larger than 1, you do this
    Question : 2x^2 + 8x +9
    1. Factorise the first two terms, but DO NOT TAKE OUT ANY X'S!
    2(x^2 + 4x) + 9

    2. Complete the square on the middle bit like before and multiply the number inside by itself and then -1 again.
    You will need 2 sets of brackets
    2((x + 2)^2 -4) +9

    3. Expand the first bracket
    The minus 8 is from the 2 x -4
    2(x + 2)^2 -8 +9

    4. And tidy tidy,

    2(x + 2)^2 + 1
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    Since you've been helped already I'll be facetious

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    (Original post by 13 1 20 8 42)
    Since you've been helped already I'll be facetious

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    Knew it'd be you

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