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Completing the square

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    I have this question.


    5x^2 – 4x – 2 = 0

    I did this,

    I divided the whole equation by 5

    x^2-0.8x-0.4=0

    Then

     (x-0.4x)^2.....=0

    I know that I'm meant to do something like move the 0.16 to the other side but I'm not too sure.

    Help
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    (Original post by zed963)
    I have this question.


    5x^2 – 4x – 2 = 0

    I did this,

    I divided the whole equation by 5

    x^2-0.8x-0.4=0

    Then

     (x-0.4x)^2.....=0

    I know that I'm meant to do something like move the 0.16 to the other side but I'm not too sure.

    Help
    When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e. x^2 - 0.8x = (x-0.4)^2 - 0.16
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    (Original post by hassi94)
    When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e. x^2 - 0.8x = (x-0.4x)^2 - 0.16
    I've been watching quite a few videos and guides, I'm getting to the point where I'm confused, I need to stick to one method that works best for me.

    The question came from this site.

    http://www.mathsisfun.com/algebra/co...ng-square.html

    But I'm not fully understanding why they do such things.
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    (Original post by hassi94)
    When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e. x^2 - 0.8x = (x-0.4x)^2 - 0.16
    it is  (x-0.4)^2 \ldots not  (x-0.4x)^2 \ldots
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    So basically  (x-0.4)^2-0.16 = 0.4 + 0.16

    If I'm correct

    I'm guessing that's wrong anyway.
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    How do I carry on this question, I'm basically stuck.
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    (Original post by zed963)
    I've been watching quite a few videos and guides, I'm getting to the point where I'm confused, I need to stick to one method that works best for me.

    The question came from this site.

    http://www.mathsisfun.com/algebra/co...ng-square.html

    But I'm not fully understanding why they do such things.
    Right well what you're doing when completing the square is adding a little bit which must be taken away again to keep the maths consistent.

    say if we want to complete the square on

    [x^2 + 2x] + 2

    I've put square brackets around the bit we'll actually be modifying the appearance of.

    So we do the regular thing here;

    = [(x+1)^2 + adj] + 2

    I have written 'adj' to indicate an adjustment that needs to be made, this is because (x+1)^2 \neq x^2 + 2x, in fact; (x+1)^2 = x^2 + 2x + 1 as we can clearly see by multiplying out.

    We are just trying to get x^2 + 2x and we don't want the '+1' so our adjustment is going to be to take away 1, i.e. adj = -1

    So now we have

    [(x+1)^2 - 1] + 2 which we then tidy up to get (x+1)^2 + 1

    since the constant term from the (x+1)^2 is 1, and similarly if it was (x+2)^2 it would be 4 etc. It is always the square of the constant term inside the bracket that is being squared.

    So, generally for a quadratic [x^2 + bx] + c, we'd write it as [(x+\frac{b}{2})^2 + adj] + c and then the adjustment would be -\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4} to give us [x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c
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    (Original post by raheem94)
    it is  (x-0.4)^2 \ldots not  (x-0.4x)^2 \ldots
    Thanks - just a typo
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    (Original post by zed963)
    How do I carry on this question, I'm basically stuck.
    Watch this video.
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    (Original post by zed963)
    I have this question.


    5x^2 – 4x – 2 = 0

    I did this,

    I divided the whole equation by 5

    x^2-0.8x-0.4=0
    Don't know about the method you are using. I'll try to explain the method I find easiest.
    If you think of your formula as x^2+px+q=0

    Carrying on from the last step I quoted, I use this formula:

    x1,2 = - p/2 ± √(p²/4 -q)

    Which gives me

    x1 = 1.15
    x2 = -0.35
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    (Original post by hassi94)
    Right well what you're doing when completing the square is adding a little bit which must be taken away again to keep the maths consistent.

    say if we want to complete the square on

    [x^2 + 2x] + 2

    I've put square brackets around the bit we'll actually be modifying the appearance of.

    So we do the regular thing here;

    = [(x+1)^2 + adj] + 2

    I have written 'adj' to indicate an adjustment that needs to be made, this is because (x+1)^2 \neq x^2 + 2x, in fact; (x+1)^2 = x^2 + 2x + 1 as we can clearly see by multiplying out.

    We are just trying to get x^2 + 2x and we don't want the '+1' so our adjustment is going to be to take away 1, i.e. adj = -1

    So now we have

    [(x+1)^2 - 1] + 2 which we then tidy up to get (x+1)^2 + 1

    since the constant term from the (x+1)^2 is 1, and similarly if it was (x+2)^2 it would be 4 etc. It is always the square of the constant term inside the bracket that is being squared.

    So, generally for a quadratic [x^2 + bx] + c, we'd write it as [(x+\frac{b}{2})^2 + adj] + c and then the adjustment would be -\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4} to give us [x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c
    I understand everything else apart from this.

    So, generally for a quadratic [x^2 + bx] + c, we'd write it as [(x+\frac{b}{2})^2 + adj] + c and then the adjustment would be -\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4} to give us [x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c

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    (Original post by qua)
    Don't know about the method you are using. I'll try to explain the method I find easiest.
    If you think of your formula as x^2+px+q=0

    Carrying on from the last step I quoted, I use this formula:

    x1,2 = - p/2 ± √(p²/4 -q)

    Which gives me

    x1 = 1.15
    x2 = -0.35

    When did you learn this?

    It seems like this method was from the 90s no offence.
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    (Original post by zed963)
    When did you learn this?

    It seems like this method was from the 90s no offence.
    And it still works wonders. :cool:

    And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways.
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    (Original post by qua)
    And it still works wonders. :cool:

    And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways.
    Oh right no wonder I wasn't familiar with it.
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    (Original post by qua)
    And it still works wonders. :cool:

    And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways.
    That is just a less general form of the quadratic equation and does not help the OP complete the square.

    (Original post by zed963)
    I understand everything else apart from this.

    So, generally for a quadratic [x^2 + bx] + c, we'd write it as [(x+\frac{b}{2})^2 + adj] + c and then the adjustment would be -\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4} to give us [x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c

    What don't you understand about it? I'm saying if we have any quadratic x^2 + bx + c we would complete the square as follows. b and c just represent constant numbers, in the example I showed you above, b = 2 and c = 2.
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    (Original post by hassi94)
    That is just a less general form of the quadratic equation and does not help the OP complete the square.



    What don't you understand about it? I'm saying if we have any quadratic x^2 + bx + c we would complete the square as follows. b and c just represent constant numbers, in the example I showed you above, b = 2 and c = 2.
    I'm just wondering where you have got the (x+b/2)^2 from
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    (Original post by zed963)
    I'm just wondering where you have got the (x+b/2)^2 from
    like when we have x^2 + 4x, we do (x+2)^2, always putting half of the x coefficient in the bracket. So if we have x^2 + bx, we'd put it as (x+b/2)^2

    Make sense?
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    (Original post by hassi94)
    like when we have x^2 + 4x, we do (x+2)^2, always putting half of the x coefficient in the bracket. So if we have x^2 + bx, we'd put it as (x+b/2)^2

    Make sense?
    Oh right I understand I thought it was something much more complicated.

    Thanks

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