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Math Help - Hi, need help with completing the square

  1. #1
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    Hi, need help with completing the square

    Hi I'm Harvey in London and I'm back at college after almost two decades of no academic studies and have my 1st assignment due in a week. I've managed to get all the questions answered but have been struggling with a particular on completing the square to solver a quadratic and have two issues with it:

    Firstly I keep getting surd form answers for my x values, is this ok if I am to sketch the curve for it???

    Secondly, that's if it's not ok, then can someone please tell me where I'm going wrong.

    I have attached the question and my method of solving in a word doc to this message,

    Please please can help me out here; I don't know how many hours I spent on this one question already

    P.S. as this is an assignment question, obviously, I'm not asking for the answers just help with methodology
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  2. #2
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    Re: Hi, need help with completing the square

    Hello, Harvey!

    Solve by completing the square: . y \;=\;1 + 3x - 72x^2

    We have: . y \;=\;\text{-}72x^2 + 3x +  1

    Factor: n . y \;=\; \text{-}72\left(x^2 - \tfrac{1}{24}x - \tfrac{1}{72}\right)

    Then: . . . y\;=\; \text{-}72\left(x^2 - \tfrac{1}{24}x {\color{red}\:+\: \tfrac{1}{2304}} - \tfrac{1}{72} {\color{red}\:-\: \tfrac{1}{2304}}\right)

    n . . . . . . y \;=\;\text{-}72\left[\left(x-\tfrac{1}{48}\right)^2 - \tfrac{33}{2304}\right]

    n . . . . . . y \;=\;\text{-}72\left(x-\tfrac{1}{48}\right)^2 + \tfrac{33}{32}

    Got it?

    Thanks from Harvey
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Hi, need help with completing the square

    I opened the document, and it appears you want to solve by completing the square:

    -2x^2+3x+1=0

    There are two methods I know of to do this. The first is the method I was taught in school:

    -2\left(x^2-\frac{3}{2}x \right)=-1

    x^2-\frac{3}{2}x=\frac{1}{2}

    x^2-\frac{3}{2}x+\left(\frac{1}{2}\cdot\frac{3}{2} \right)^2=\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{  3}{2} \right)^2

    \left(x-\frac{3}{4} \right)^2=\frac{17}{16}

    x-\frac{3}{4}=\pm\frac{\sqrt{17}}{4}

    x=\frac{3\pm\sqrt{17}}{4}

    The second method I picked up on another forum:

    -2x^2+3x+1=0

    2x^2-3x=1

    Multiply through by 4(2)=8

    16x^2-24x=8

    Add (-3)^2=9 to both sides:

    16x^2-24x+9=17

    (4x-3)^2=17

    4x-3=\pm\sqrt{17}

    x=\frac{3\pm\sqrt{17}}{4}
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  4. #4
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    Re: Hi, need help with completing the square

    Thanks ever so much guys, you're both great people!! It turns out that my problem with the surds wasn't a problem at all and that in order to treat them as intercepts of the parabola on x-axis I just need to convert to decimal forms, duh. Thanks a lot for caring enough to post a reply and it's brilliant to have another way of completing the square, I shall show that to my maths teacher.

    Keep calculating

    Cheers
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