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Math Help - Completing the square

  1. #1
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    Completing the square

    Hello,

    I'm having problems with completing the square can anyone comment on my working?

    2x^2 - 3x -1 = 0

    2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

    2(x - 3/4)^2 -17/16 =0

    2(x - 3/4)^2 = 17/16

    (x - 3/4)^2 = 17/32

    x - 3/4 = +-√17/32

    x = 3/4 +-√17/32

    Thanks in advance. Please post solution if I am incorrect.
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  2. #2
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    Quote Originally Posted by Mouseman View Post
    Hello,

    I'm having problems with completing the square can anyone comment on my working?

    2x^2 - 3x -1 = 0

    2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

    2(x - 3/4)^2 -17/16 =0

    2(x - 3/4)^2 = 17/16

    (x - 3/4)^2 = 17/32

    x - 3/4 = +-√17/32

    x = 3/4 +-√17/32

    Thanks in advance. Please post solution if I am incorrect.
    2x^2 - 3x = 1

    x^2 - (3/2)x = 1/2

    (x - 3/4)^2 - 9/16 = 1/2

    x - 3/4 = +-sqrt(17/16)

    x = 3/4 +-sqrt(17/16)
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  3. #3
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    Thank you very much!
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  4. #4
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    Quote Originally Posted by sean.1986 View Post
    2x^2 - 3x = 1

    x^2 - (3/2)x = 1/2

    (x - 3/4)^2 - 9/16 = 1/2

    x - 3/4 = +-sqrt(17/16)

    x = 3/4 +-sqrt(17/16)
    The answer is to be x = 3/4 +- √17/4

    I can't see where I am going wrong.
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  5. #5
    Eater of Worlds
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    When completing the square, just use the general formula.

    a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}

    Plug in a,b,c and you're set.
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  6. #6
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    Second attempt

    2x^2 - 3x -1 = 0

    2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

    2((x - 3/4)^2) -17/8 =0

    2(x - 3/4)^2 = 17/8

    (x - 3/4)^2 = 17/16

    x - 3/4 = +-√17/16

    x = 3/4 +-√17/16

    Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
    Last edited by Mouseman; May 21st 2008 at 06:27 AM. Reason: Added text
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  7. #7
    A riddle wrapped in an enigma
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    Just to add my 2 cents......


    2x^2-3x-1=0

    1. First transpose the -1 to the right side of the equation.

    2x^2-3x=1

    2. Divide each term by 2

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

    3. Take half of the coefficient of x, square it and add it to both sides.

    x^2-\frac{3}{2}x+(\frac{3}{4})^2=\frac{1}{2}+\frac{9}{  16}

    4. Noting the perfect square trinomial on the left:

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

    5. Take the square root of both sides:

    x-\frac{3}{4}=\pm\sqrt\frac{17}{16}

    6. Finally,

    x=\frac{3}{4}\pm\frac{\sqrt17}{4}

    x=\frac{3\pm\sqrt17}{4}
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  8. #8
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    Hello,

    Quote Originally Posted by Mouseman View Post
    Second attempt

    2x^2 - 3x -1 = 0

    2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0

    2((x - 3/4)^2) -17/8 =0

    2(x - 3/4)^2 = 17/8

    (x - 3/4)^2 = 17/16

    x - 3/4 = +-√17/16

    x = 3/4 +-√17/16

    Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
    Note that 16=4^2

    Therefore, \sqrt{16}=4, and this yields the result your teacher gave you
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  9. #9
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    Yeah it's just a question of bracketing or simplifying.

    sqrt (a/b) = sqrt(a) / sqrt(b)

    so sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4

    Both answers are correct but I guess I should've simplified. Sorry mate!
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