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    x^2 - 3x + 5 in the form (x - p)^2 + q
    can anyone plx explain how do i do completing the square because i dont know if there is a method to do them
    i suck at maths lol
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    (Original post by coolmathametics)
    x^2 - 3x + 5 in the form (x - p)^2 + q
    can anyone plx explain how do i do completing the square because i dont know if there is a method to do them
    i suck at maths lol
     \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2}  + \frac{c}{a} \right)
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    (Original post by DylanJ42)
     \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + c \right)
    oml that looks way confusing lol
    would u be able to make up an example n substitute the values plx so that way i know what to substitute n what is actually happening
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    (Original post by coolmathametics)
    x^2 - 3x + 5 in the form (x - p)^2 + q
    can anyone plx explain how do i do completing the square because i dont know if there is a method to do them
    i suck at maths lol
    I recommend you watch a few videos on the topic and then come back to us if you're still stuck. You could try this video for example.
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    (Original post by DylanJ42)
     \displaystyle ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a}) = a \left( (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2}  + \frac{c}{a} \right)
    Is that really going to help someone who can't complete the square for simple quadratic expressions?
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    (Original post by coolmathametics)
    oml that looks way confusing lol
    would u be able to make up an example n substitute the values plx so that way i know what to substitute n what is actually happening
    yea sure, so something like  \displaystyle x^2 + 4x + 7

    to complete the square we write this as  \displaystyle (x + 2)^2 -4 + 7

    this is because if you multiply out the squared bracket you get  \displaystyle x^2 + 2x + 2x + 4 = x^2 + 4x + 4 , however we only want  \displaystyle x^2 + 4x so we need to subtract the 4
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    (Original post by notnek)
    I recommend you watch a few videos on the topic and then come back to us if you're still stuck. You could try this video for example.
    i can do those ones but if the middle term is 7 or 11 then its a bit difficult

    (Original post by notnek)
    Is that really going to help someone who can't complete the square for simple quadratic expressions?
    lol i did i say i suck at maths but not that much, i can do simple ones but difficulty with the ones i explained above
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    (Original post by DylanJ42)
    yea sure, so something like  \displaystyle x^2 + 4x + 7

    to complete the square we write this as  \displaystyle (x + 2)^2 -4 + 7

    this is because if you multiply out the squared bracket you get  \displaystyle x^2 + 2x + 2x + 4 = x^2 + 4x + 4 , however we only want  \displaystyle x^2 + 4x so we need to subtract the 4
    what is the middle term is 11 instead
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    (Original post by coolmathametics)
    what is the middle term is 11 instead
    are you comfortable with the idea of comparing coefficients

    eg if  \displaystyle 3x^2 + 4x \equiv  ax^2 +bx then a=3 and b=4

    does that make sense?
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    can some1 help me plx
    sorry if i seems desperate
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    (Original post by coolmathametics)
    i can do those ones but if the middle term is 7 or 11 then its a bit difficult
    You should have mentioned that in your initial post

    Can you try completing the square for x^2 - 3x + 5?

    It will be useful if you show us where you get stuck so please post all your working / thoughts up to the point you get stuck.

    Also, is this for A Level or GCSE? Will you be using a calculator to do these types of questions?
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    (Original post by coolmathametics)
    can some1 help me plx
    sorry if i seems desperate
    If you have x^2+bx+c then it can be written in the form (x+\frac{b}{2})^2-(\frac{b}{2})^2+c

    So in your question you have b=-3 and c=5 - all it means is that an odd number being halved will give you a fraction to work with instead.

    To understand why this works, you should watch a video on it and see how this is derived.
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    (Original post by notnek)
    You should have mentioned that in your initial post

    Can you try completing the square for x^2 - 3x + 5?

    It will be useful if you show us where you get stuck so please post all your working / thoughts up to the point you get stuck.

    Also, is this for A Level or GCSE? Will you be using a calculator to do these types of questions?
    ooops sorry but i can do e.g. x^2 + 4x + 6
    because i k 2 + 2 would be 4
    so its gonna be (x+2)^2 + 2 i think :/
    regards to ur question, i dunno how to start like the first step cox i can't find the half of 3x so its gonna be like 3/2 which is gonna be difficult to see how to work out the part that goes outside of the bracket... if u see what i mean

    its for A level - core 1 non cal
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    (Original post by DylanJ42)
    are you comfortable with the idea of comparing coefficients

    eg if  \displaystyle 3x^2 + 4x \equiv ax^2 +bx then a=3 and b=4

    does that make sense?
    i am so confused
    ok so lets say ---> x^2 + 7x - 3
    what i need to know is how i get half of 7 cox it cant be divided by 2
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    (Original post by coolmathametics)
    x^2 - 3x + 5 in the form (x - p)^2 + q
    can anyone plx explain how do i do completing the square because i dont know if there is a method to do them
    i suck at maths lol
    http://www.mathsgenie.co.uk/completing-the-square.html
    Watch this ^ the guy explains it quickly and easily.
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    (Original post by coolmathametics)
    i am so confused
    ok so lets say ---> x^2 + 7x - 3
    what i need to know is how i get half of 7 cox it cant be divided by 2
    its just  \displaystyle \frac{7}{2} , fractions are allowed in completing the square
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    (Original post by DylanJ42)
    its just  \displaystyle \frac{7}{2} , fractions are allowed in completing the square
    alright but how am i gonna work out + q (the outside bit)
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    (Original post by Anonymous1502)
    http://www.mathsgenie.co.uk/completing-the-square.html
    Watch this ^ the guy explains it quickly and easily.
    i can do those ones pretty easily cox there ain't any fractions in the answer *_*
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    (Original post by coolmathametics)
    ooops sorry but i can do e.g. x^2 + 4x + 6
    because i k 2 + 2 would be 4
    so its gonna be (x+2)^2 + 2 i think :/
    regards to ur question, i dunno how to start like the first step cox i can't find the half of 3x so its gonna be like 3/2 which is gonna be difficult to see how to work out the part that goes outside of the bracket... if u see what i mean

    its for A level - core 1 non cal
    There are different ways to think about 'completing the square' but I recommend this method when there are fractions and it's non calculator:

    E.g. x^2 - 3x + 5

    Start by halving -3 to get -\frac{3}{2} as you say

    So you get \left(x-\frac{3}{2}\right)^2

    Then you need to subtract the square of -\frac{3}{2}\right and then add on 5 (let me know if you can't see where I'm getting these numbers from).

    So you end up with :

    \displaystyle \left(x-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 + 5

    \displaystyle =\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + 5


    Now change the 5 so it has the same denominator as \frac{9}{4}

    \displaystyle =\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + \frac{20}{4}

    \displaystyle =\left(x-\frac{3}{2}\right)^2+\frac{11}{4  }


    Now see if you can follow this process for a different example. Post your working if you get stuck.
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    (Original post by coolmathametics)
    alright but how am i gonna work out + q (the outside bit)
    as notnek said, try do an example so we can see where you get stuck.

    try complete the square on  \displaystyle x^2 - 3x + 5
 
 
 
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