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I can't find the difference of two squares. I don't understand :( Watch

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    i can do simple ones like :2x^2+6x
    but struggling to do ones like
    2x^2+5x+2

    3x^2+10x−8

    15x^2+42x−9


    Would be great if someone could go through how you solve them.
    Thanks x
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    (Original post by LikeAStar)
    i can do simple ones like :2x^2+6x
    but struggling to do ones like
    2x^2+5x+2

    3x^2+10x−8

    15x^2+42x−9


    Would be great if someone could go through how you solve them.
    Thanks x
    Those aren't differences between two squares.
    A difference between two sqaures has an X^2 term and a number term which is a square number.
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    2x^2 + 5x +2 i think you just complete the square..... and the following 2 as well... can use the quadratic formula as well.
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    you mean you can't complete the square?
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    Looks as though you have to factorise those expressions.
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    complete the square, factorise or use the quadratic formula.
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    i never really learnt to complete the square, its a pretty useless skill. factorising and using the quadratic formula are far more usefull.
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    got it now.
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    (Original post by LikeAStar)
    i can do simple ones like :2x^2+6x
    but struggling to do ones like
    2x^2+5x+2

    3x^2+10x−8

    15x^2+42x−9


    Would be great if someone could go through how you solve them.
    Thanks x
    Those are not difference of two squares.

    I'll take it you mean factorising? Or completing the square? I'll do both.

    For factorising you need to come up with your own method. If you really have no iea how to factorise, PM me and I'll teach you my method that always works. Here are the answers anyway

    Spoiler:
    Show

    2x^2+5x+2 = (2x+1)(x+2)

    3x^2+10x-8=(3x-2)(x+4)

    15x^2+42x-9 = 3(5x^2+14x-3) = 3(5x-1)(x+3)


    For completing the square, here is the process:

    - Factorise everything through by the number in front of the x^2 ( "a" )
    2x^2+5x+2 = 2[x^2+\frac{5}{2}x+1]

    - Divide the co-eff of the x-term (we'll call this co-eff "b" ) by two, then put the first two terms into brackets as (x + b/2)^2, then SUBTRACT ((b/2)^2). Don't forget to put the constant "c" down to the next line. This may sound complicated, but it's really not.
    2[(x+\frac{5}{4})^2-(\frac{5}{4})^2+1]

    - Simply what's inside the square brackets
    2[(x+\frac{5}{4})^2-(\frac{25}{16})+1]=2[(x+\frac{5}{4})^2-\frac{9}{16}]

    - Multiply out by what's in front of the square brackets
    2(x+\frac{5}{4})^2-\frac{9}{8}

    And that's your final answer!
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    (Original post by didgeridoo12uk)
    i never really learnt to complete the square, its a pretty useless skill. factorising and using the quadratic formula are far more usefull.
    You are not serious, are you?
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    (Original post by steve2005)
    You are not serious, are you?
    i'm totally serious,

    you can use it to solve a quadratic, and find the turning point.

    but in nearly all cases its easier to factorise/use the formula, and differentiate to get exactly the same information. and you're less likely to make mistakes
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    (Original post by didgeridoo12uk)
    i'm totally serious, what possible use is completing the square?
    How about translations of quadratic graphs for one?
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    (Original post by milliondollarcorpse)
    How about translations of quadratic graphs for one?
    that can be done in other more general ways, that will work for practically all graphs not just quadratics.


    its just got far too specific an application to be of any real use. nobody would ever use it at uni
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    (Original post by didgeridoo12uk)
    i'm totally serious,

    you can use it to solve a quadratic, and find the turning point.

    but in nearly all cases its easier to factorise/use the formula, and differentiate to get exactly the same information. and you're less likely to make mistakes
    I thought you were a student at Cambridge, however, I now think you are not at university studying mathematics. IF you are then I think something is not right with the state of education in the UK
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    (Original post by steve2005)
    I thought you were a student at Cambridge, however, I now think you are not at university studying mathematics. IF you are then I think something is not right with the state of education in the UK
    studying engineering, not maths.

    and you've still failed to show me the point of it.


    it's like the people the memorise every single formula that could be of any use. rather than just learning the main ones, and how to derive the rest from first principles
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    (Original post by didgeridoo12uk)
    studying engineering, not maths.

    and you've still failed to show me the point of it.


    it's like the people the memorise every single formula that could be of any use. rather than just learning the main ones, and how to derive the rest from first principles
    You were the one that said that completing the square was not necessary or useful and not worth learning. I point out that it is covered in GCSE.

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    (Original post by steve2005)
    You were the one that said that completing the square was not necessary or useful and not worth learning. I point out that it is covered in GCSE.

    i'd personally multiply out the equation of a circle

    (x-a)^2 + (y-b)^2 = r^2

    and compare coefficients.

    might take slightly longer, but its one less thing i'd have to remember for an exam.
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    (Original post by didgeridoo12uk)
    i'd personally multiply out the equation of a circle

    (x-a)^2 + (y-b)^2 = r^2

    and compare coefficients.

    might take slightly longer, but its one less thing i'd have to remember for an exam.
    I see no difference between doing that and completing the square. The important part of completing the square is knowing that you can rewrite x^2+bx+c in the form (x+B)^2+C for some B, C. Knowing the actual formulae for B and C is unimportant, but just because you don't bother to remember them doesn't mean you're not completing the square.
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    Completing the square can also be used for finding the minimum/maximum points for a Quadratic. Not sure now if there is an alternative method.
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    (Original post by didgeridoo12uk)
    i'd personally multiply out the equation of a circle

    (x-a)^2 + (y-b)^2 = r^2

    and compare coefficients.

    might take slightly longer, but its one less thing i'd have to remember for an exam.
    All you need to 'remember' is that you're putting the quadratic into the form of a perfect square (with a constant outside). There's no formula or anything.

    (Original post by Liverpool F.C.™)
    Completing the square can also be used for finding the minimum/maximum points for a Quadratic. Not sure now if there is an alternative method.
    Differentiate?
 
 
 
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