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    Give your solutions to 2 decimal places.
    I'm guessing you start of like this:
    (x )(x )
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    (Original post by multiplexing-gamer)
    Give your solutions to 2 decimal places.
    I'm guessing you start of like this:
    (x )(x )
    Use the quadratic formula
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    (Original post by multiplexing-gamer)
    Give your solutions to 2 decimal places.
    I'm guessing you start of like this:
    (x )(x )
    Use the quadratic formula
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    try the formula
    http://www.google.co.uk/imgres?imgur...1t:429,r:2,s:0
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    where a - number in front of x^2, b is number infront of x and c is the last number
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    complete the square
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    Oh yeh thanks guys
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    quadratic formula, it will be at the fron of the paper, but you have to give the + and the - values
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    (Original post by multiplexing-gamer)
    Give your solutions to 2 decimal places.
    I'm guessing you start of like this:
    (x )(x )
    The "solutions to 2dp" gives you a clue that it won't factorise with integer values. You can complete the square or use the quadratic formula. Since they don't want an exact form, putting everything into the quadratic formula will probably be faster.
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    (Original post by multiplexing-gamer)
    Give your solutions to 2 decimal places.
    I'm guessing you start of like this:
    (x )(x )
    Use x=\frac {-b \pm \sqrt{b^2-4ac}} {2a}

    a = the number in front of the x^2 = 1
    b= -2 (the number in front of the x) so -b = 2
    c= -1 ( the last number)

    the plus-minus sign means that there are two solutions (use the plus once and minus once)

    The mathematical term for 'in front of' is coefficient btw.
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    (x-1)^2 -2 = 0
    (x-1)^2 =2
    x-1 = sqrt(2)

    so x is 1 +/- sqrt(2)
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    (Original post by CuthbertDibble)
    (x-1)^2 -2 = 0
    (x-1)^2 =2
    x+1 = sqrt(2)

    so x is -1 +/- sqrt(2)

    so 0.41 and -2.41
    You made a mistake here. Your minus has changed to a plus.
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    (Original post by Gemini92)
    You made a mistake here. Your minus has changed to a plus.
    thanks
 
 
 

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