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    Hi, can someone show me a step by step method for completing the square for

    9x^2 - 12x - 2 = 0?

    What I did:

    3(3x - 4)^2 - 2 = 0
    3(3x - 2)^2 = 6
    3(3x - 2) = √6
    3x - 2 = √6 / 3
    3x = 2 + √6 / 3 (all over 3)
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    I forgot to say I need to solve for x. The answer is meant to be
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    and what I got in the first post is wrong because I have 3x instead of x =.
    This is the question
    Attachment 529103529105
    This is the whole mark scheme for part (c)

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    Thank you.
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    (Original post by JezDayy)
    Hi, can someone show me a step by step method for completing the square for

    9x2 - 12x - 2 = 0?

    What I did:

    3(3x - 4)^2 - 2 = 0
    3(3x - 2)^2 = 6
    3(3x - 2) = √6
    3x - 2 = √6 / 3
    3x = 2 + √6 / 3 (all over 3)
    write a(bx + c)2 - d = 0

    expand this & compare to

    9x2 - 12x - 2 = 0

    to find a,b,c,d
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    (Original post by the bear)
    if you multiply out the highlighted expression you get 27x2
    Oh yeah I can see that now. Would you take 9 out as a factor before completing the whole thing? I tried that too but got root 22 or something instead of root 6.
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    (Original post by JezDayy)
    Hi, can someone show me a step by step method for completing the square for

    9x^2 - 12x - 2 = 0?

    What I did:

    3(3x - 4)^2 - 2 = 0
    3(3x - 2)^2 = 6
    3(3x - 2) = √6
    3x - 2 = √6 / 3
    3x = 2 + √6 / 3 (all over 3)
    you must take out 9 as a common factor first(even if it's not common xD

    9\left(x^2 - \dfrac{4}{3} x - \dfrac{2}{9}\right)=0

    then don't forget since 9 times by everything you do this??

    9\left( \left(x-\dfrac{2}{3}\right)^2 + \dfrac{4}{3} - \dfrac{2}{9}\right)=0

    y'know
    \dfrac{4}{3} - \dfrac{2}{9}

    9\left( \left(x-\dfrac{2}{3}\right)^2 + \dfrac{10}{9}\right)=0

    then expand outer brackets to get

    9\left(x-\dfrac{2}{3}\right)^2 + 10=0

    ^^ i think that's right
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    You are supposed to use the quadratic formula for c.
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    (Original post by thefatone)
    you must take out 9 as a common factor first(even if it's not common xD9\left(x^2 - \dfrac{4}{3} x - \dfrac{2}{9}\right)=0
    Oh ok, thank you! Looks nice to work with lol

    (Original post by greentron6)
    You are supposed to use the quadratic formula for c.
    I know, I realised after looking at the answer, but I was just wondering how to complete the square in general since I couldn't do it and need to work on it.
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    (Original post by JezDayy)
    Hi, can someone show me a step by step method for completing the square for

    9x^2 - 12x - 2 = 0?

    What I did:

    3(3x - 4)^2 - 2 = 0
    3(3x - 2)^2 = 6
    3(3x - 2) = √6
    3x - 2 = √6 / 3
    3x = 2 + √6 / 3 (all over 3)
    http://www.examsolutions.net/maths-r...tutorial-1.php

    Enjoy!
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    (Original post by JezDayy)
    Oh ok, thank you! Looks nice to work with lol


    I know, I realised after looking at the answer, but I was just wondering how to complete the square in general since I couldn't do it and need to work on it.
    i edited it

    this one's a tougher one..... since it's got fractions and you took out a common factor of 9....
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    haha cheers for the link, but I actually watched this before posting here since I used his method and still got it wrong, and had no idea where I went wrong lol

    (Original post by thefatone)
    i edited it

    this one's a tougher one..... since it's got fractions and you took out a common factor of 9....
    Thank you! Really appreciate the help. If I knew the quadratic formula was an easier way I would've used it lol, I usually never use it in C1.
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    Hope this helps! Let me know if any of it doesn't make sense!

    Posted from TSR Mobile
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    (Original post by JezDayy)
    haha cheers for the link, but I actually watched this before posting here since I used his method and still got it wrong, and had no idea where I went wrong lol



    Thank you! Really appreciate the help. If I knew the quadratic formula was an easier way I would've used it lol, I usually never use it in C1.
    really????

    well..... as long as you know how to both ^-^
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    (Original post by JezDayy)
    Hi, can someone show me a step by step method for completing the square for

    9x^2 - 12x - 2 = 0?
    see attached file
    Attached Images
  1. File Type: pdf step by step how to complete...pdf (360.7 KB, 39 views)
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    (Original post by megan-garner6)
    Hope this helps! Let me know if any of it doesn't make sense!Posted from TSR Mobile
    Thank you! Really appreciate this. Is there a rule you need to remember for the root all 2 / 3 becoming root 6 / 3? I don't understand that bit.
    (Original post by thefatone)
    really????well..... as long as you know how to both ^-^
    Yeah, since it's non-calculator and numbers are usually big or the x values are whole lol

    (Original post by depymak)
    see attached file
    Thank you!
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    (Original post by JezDayy)
    Thank you! Really appreciate this. Is there a rule you need to remember for the root all 2 / 3 becoming root 6 / 3? I don't understand that bit.Yeah, since it's non-calculator and numbers are usually big or the x values are whole lol



    Thank you!
    Haha I used my calculator for that bit! I think you can leave root(2/3) as root(6/9) then just root the 9 to make 3 and leave the root on top of the 6.
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    (Original post by megan-garner6)
    Haha I used my calculator for that bit! I think you can leave root(2/3) as root(6/9) then just root the 9 to make 3 and leave the root on top of the 6.
    Oh lol, I can't use a calculator in that question!!
    I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

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    (Original post by JezDayy)
    Oh lol, I can't use a calculator in that question!!
    I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

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    Ooh that's good! I think I'll use that!
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    (Original post by JezDayy)
    Oh lol, I can't use a calculator in that question!!
    I've noticed and made up some kind of rule though lol, if you times the numerator and the denominator of the fraction, put the number you get on the numerator and square root that only, then put that over the original denominator. I need to remember this method haha

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    LOL some weird rule eh?
    that's called rationalising the denominator, i.e multiply by something to get rid of the surd on the bottom

    let's take a look

    \sqrt {\dfrac{2}{5}} = \dfrac{\sqrt 2}{\sqrt 5}

    multiply top and bottom by  \sqrt 5 to get rid of the  \sqrt 5on the bottom right?

     \dfrac{\sqrt 2}{ \sqrt 5} \times \dfrac{\sqrt 5}{\sqrt 5}

     \dfrac{\sqrt 5 \times \sqrt 2}{\sqrt 5 \times \sqrt 5}

     \dfrac{\sqrt {5 \times 2}}{\sqrt {5\times 5}}

    everything cancels and works out as

     \dfrac{\sqrt 10}{5}
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    (Original post by thefatone)
    LOL some weird rule eh?
    that's called rationalising the denominator, i.e multiply by something to get rid of the surd on the bottom

    let's take a look

    \sqrt {\dfrac{2}{5}} = \dfrac{\sqrt 2}{\sqrt 5}

    multiply top and bottom by  \sqrt 5 to get rid of the  \sqrt 5on the bottom right?

     \dfrac{\sqrt 2}{ \sqrt 5} \times \dfrac{\sqrt 5}{\sqrt 5}
    OH right LOL I know what rationalising the denominator is, I just didn't know you could do it there! I don't think I'm prepared for C1 at all lol
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    (Original post by JezDayy)
    OH right LOL I know what rationalising the denominator is, I just didn't know you could do it there! I don't think I'm prepared for C1 at all lol
    LOL it's ok, i edited it and finished it off xD
 
 
 
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