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    Find, as surds, the roots of the equation:

    2(x+1)(x-4) - (x-2)^2=0

    Whats the answer?
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    (Original post by Chrisrc)
    Find, as surds, the roots of the equation:

    2(x+1)(x-4) - (x-2)^2=0

    Whats the answer?
    This isn't a homework service.

    Expand the brackets (be careful with signs when performing the subtraction) then solve the resulting quadratic by completing the square or by using the quadratic formula.
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    I got 5+- square root 29 but my book says it's 1+- square root 13, I can't see how it got that answer?
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    (Original post by Chrisrc)
    I got 5+- square root 29 but my book says it's 1+- square root 13, I can't see how it got that answer?
    The book is right. What quadratic did you get?
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    I tried it with both 'completing the square' and 'ax^2 + bx + c" formula and got the same answer
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    (Original post by Chrisrc)
    I tried it with both 'completing the square' and 'ax^2 + bx + c" formula and got the same answer
    I'll ask again. What was the quadratic you were trying to solve?
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    2(x+1)(x-4) - (x-2)^2 = 0
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    (Original post by Chrisrc)
    2(x+1)(x-4) - (x-2)^2 = 0
    Blimey. This is hard work. What did you get after you expanded the brackets and collected like terms?
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    Oooh sarry, I got x^2 -10x - 4 = 0
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    (Original post by Chrisrc)
    Oooh sarry, I got x^2 -10x - 4 = 0
    It should be x^2-2x-12=0

    If you look back to my first post I told you to be careful when subtracting. You weren't careful. There are subtractions of negative numbers to deal with.
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    What do you get when you expand
    2(x+1)(x-4) and when you expand -(x-2)^2
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    (Original post by Chrisrc)
    What do you get when you expand
    2(x+1)(x-4) and when you expand -(x-2)^2
    2(x^2 -3x-4)-(x^2-4x+4)=0
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    Oh I see I got that too but I didn't put the x^2 - 4x + 4 in brackets and times it by -1... Thanks for your help
 
 
 
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