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    Solve 2x^2+x-351=0
    By the way, the only way I want this presented is by completing the square.
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    Is this a way of getting us to do your homework?
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x+351=0
    By the way, the only way I want this presented is by completing the square.
    divide through by 2 and then do completing the square as usual ?
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x+351=0
    By the way, the only way I want this presented is by completing the square.
    Yes, I can solve it. I would, however, prefer for you to try first.

    Just so you know, both roots are complex, so they are far beyond GCSE understanding.
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    (Original post by TheOtherSide.)
    Is this a way of getting us to do your homework?
    No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.
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    (Original post by Ioniccopperflame)
    No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.
    You shouldn't be getting real solutions, unless you've made an error in copying the question.
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    (Original post by Ioniccopperflame)
    No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.
    Neither of those is a solution to the equation that you posted - was there an error in it? Should it have been -351?
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x+351=0
    By the way, the only way I want this presented is by completing the square.
    x=+- 69, for all real and imaginary values of x in da world :O
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    (Original post by _gcx)
    Yes, I can solve it. I would, however, prefer for you to try first.

    Just so you know, both roots are complex, so they are far beyond GCSE understanding.
    No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not when I tried solving it through completing the square.
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    (Original post by Ioniccopperflame)
    No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not when I tried solving it through completing the square.
    Both of those are wrong. This equations has no real roots.
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x-351=0
    By the way, the only way I want this presented is by completing the square.
    I'm so sorry. I realised I made a mistake in the forum. I meant to type -351 instead of +351 but it's changed now
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x-351=0
    By the way, the only way I want this presented is by completing the square.
    I'm so sorry. I realised I made a mistake in the forum. I meant to put -351 instead of +351. I've changed it now.
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    (Original post by Ioniccopperflame)
    Solve 2x^2+x-351=0
    By the way, the only way I want this presented is by completing the square.
    I'll try 2x^2+x+351 for fun anyway
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     \displaystyle x^2+\frac{1}{2}x-\frac{351}{2}=0
     \displaystyle \left (x+\frac{1}{4}\right )^2-\frac{1}{16}-\frac{2808}{16}=0
     \displaystyle \left (x+\frac{1}{4}\right )^2 =\frac{2809}{16} .
    I'm assuming you can do it from here.
 
 
 
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