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    It says to find the exact solutions of x^2 - 6x - 3 = 0 by completing the square, but I can't find the exact solutions. All I get is x = 3 + root 12 and x = 3 - root 12

    Then it says to sketch the graph of y = (2x-13)(x^2-6x-3) but I can't unless I find the exact solutions because I need to factorise the above equation into three linear factors. help please!
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    (Original post by vedderfan94)
    x = 3 + root 12 and x = 3 - root 12
    Correct, but you could simplify your surds.

    This is what they mean by exact values.
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    (Original post by Get me off the £\?%!^@ computer)
    Correct, but you could simplify your surds.

    This is what they mean by exact values.
    Ok so I get x = 3 + or - 2root3 but then what? how do I sketch the graph of the cubic in my first post?
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    If x = 3 + 2 \sqrt{3} then x - 3 - 2 \sqrt{3} = 0

    You can use this to write down a linear factor.

    You don't need factors in order to sketch the graph anyway, it is solutions you need.
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    Exact solution means to leave it in surd form!
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    (Original post by Mr M)
    If x = 3 + 2 \sqrt{3} then x - 3 - 2 \sqrt{3} = 0

    You can use this to write down a linear factor.

    You don't need factors in order to sketch the graph anyway, it is solutions you need.
    I still don't get it. How do i use x -3 -2root 3 =0 to find the linear factors.
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    (Original post by vedderfan94)
    I still don't get it. How do i use x -3 -2root 3 =0 to find the linear factors.
    (x-root)(x- another root)( x - another root)=0
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    (Original post by vedderfan94)
    I still don't get it. How do i use x -3 -2root 3 =0 to find the linear factors.
    If a quadratic has x = k as a solution then x - k = 0 and (x - k) is a linear factor. Understand now?

    Anyway, as I said, you don't need factors to sketch the graph. You need the solutions and you already have them.
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    (Original post by vedderfan94)
    It says to find the exact solutions of x^2 - 6x - 3 = 0 by completing the square, but I can't find the exact solutions. All I get is x = 3 + root 12 and x = 3 - root 12

    Then it says to sketch the graph of y = (2x-13)(x^2-6x-3) but I can't unless I find the exact solutions because I need to factorise the above equation into three linear factors. help please!
    In order to sketch the graph, one of the things you are working out is solving the equation for when y=0. So you are looking for the solutions to

    (2x-13)(x^2-6x-3)=0

    Well if the product of two terms is zero, then one or other of the two terms must be zero, so either (2x-3) = 0 or (x^2-6x-3)=0

    The first one is easy to solve, and you have already solved the second one, hence you have your three points where the curve crosses the x axis, i.e. where y=0.
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    (Original post by ghostwalker)
    In order to sketch the graph, one of the things you are working out is solving the equation for when y=0. So you are looking for the solutions to

    (2x-13)(x^2-6x-3)=0

    Well if the product of two terms is zero, then one or other of the two terms must be zero, so either (2x-3) = 0 or (x^2-6x-3)=0

    The first one is easy to solve, and you have already solved the second one, hence you have your three points where the curve crosses the x axis, i.e. where y=0.
    I don't get the second bracket though. Where does x = 3 + 2root3 cross the x-axis?
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    (Original post by vedderfan94)
    I don't get the second bracket though. Where does x = 3 + 2root3 cross the x-axis?
    Type \sqrt 3 into your calculator and you will get, roughly, 1.7. Then put that into  3+2\pm\sqrt3 and you will get the approximate roots of the second brackets which will help position them relative to the other root you know. Then label these roots on the x axis
    (Its just a sketch so doesn't need to be exact) in the exact form and then do the same for the y axis (just put x to zero).
    Its a positive x^3 graph so its just a standard s shape.
    If you want more detail you will have to look at the second derivative.
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    (Original post by ben-smith)
    Type \sqrt 3 into your calculator and you will get, roughly, 1.7. Then put that into  3+2\pm\sqrt3 and you will get the approximate roots of the second brackets which will help position them relative to the other root you know. Then label these roots on the x axis
    (Its just a sketch so doesn't need to be exact) in the exact form and then do the same for the y axis (just put x to zero).
    Its a positive x^3 graph so its just a standard s shape.
    If you want more detail you will have to look at the second derivative.
    Thanks for the help. I understand it now
 
 
 
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