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    I am stuck on part c of this question. Can you please help me?? A step by step explanation would be useful!

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    if you have

    x(a quadratic in x) = 0

    then either x = 0 or the quadratic = 0

    since they want surds in the answer it is not likely that the quadratic will factorise.
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    (Original post by the bear)
    if you have

    x(a quadratic in x) = 0

    then either x = 0 or the quadratic = 0

    since they want surds in the answer it is not likely that the quadratic will factorise.
    how am i supposed to find the coordinates?
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    You need to start by finding the roots of the equation x(x^2-8x+4). This can be found by using the quadratic formula on the inside part of the bracket. This will then give you three answers: 0 (because of the x outside of the brackets) and two surds. The positive one is the x-coordinate.

    Next, you need to put the x-coordinate into one of the original formulae in order to work out the y-coordinate.

    Hope this helps!
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    (Original post by ayylmao97)
    how am i supposed to find the coordinates?
    you get the x coordinates by solving the equation, then to find the y coordinates you put the x coordinates into one of the two original functions ( it does not matter which ).
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    Did you get the answer?
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    (Original post by Student 977)
    Did you get the answer?
    no, i don't know how to solve the equation
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    (Original post by ayylmao97)
    no, i don't know how to solve the equation
    You need to use the quadratic formula on (x^2-8x+4). This will give you two answers as surds. Can you do that?
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    (Original post by Student 977)
    You need to use the quadratic formula on (x^2-8x+4). This will give you two answers as surds. Can you do that?
    so far i got to (8+-√48)/2 but i don't know how to cancel down the root
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    (Original post by ayylmao97)
    so far i got to (8+-√48)/2 but i don't know how to cancel down the root
    I forgot it was a non-calculator paper! You need to complete the square of x^2-8x+4.
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    Name:  Maths.jpg
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    Here is some help!
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    (Original post by Student 977)
    I forgot it was a non-calculator paper! You need to complete the square of x^2-8x+4.
    I've done it, and I got (x-4)^2-12
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    (Original post by ayylmao97)
    I've done it, and I got (x-4)^2-12
    Yes and then you can equate that to x
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    (Original post by Student 977)
    Yes and then you can equate that to x
    After that do I substitute the x coordinate into one of the equations to find y?
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    (Original post by ayylmao97)
    After that do I substitute the x coordinate into one of the equations to find y?
    Yes, but make sure you use the correct x-coordinate.
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    (Original post by Student 977)
    Yes, but make sure you use the correct x-coordinate.
    Lool sorry to keep bothering you like this but I don't know how to expand the brackets once I've substituted x
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    Which equation did you put the x into?
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    (Original post by Student 977)
    Which equation did you put the x into?
    Into the x(4-x)
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    So you get (4-2rt3)(4-4-2rt3)
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    (Original post by Student 977)
    So you get (4-2rt3)(4-4-2rt3)
    yes but then wouldnt the second bracket cancel down to (4-2rt3)?
 
 
 
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