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c1 question help

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    Hi can someone help me, I would of had an attempt but I have no idea where to begin. Is this discriminant question? Help would be appreciated
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    (Original post by junayd1998)
    Hi can someone help me, I would of had an attempt but I have no idea where to begin. Is this discriminant question? Help would be appreciated
    Start by equating the two equations, they don't intersect, so the quadratic would have no real solutions.
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    (Original post by NotNotBatman)
    Start by equating the two equations, they don't intersect, so the quadratic would have no real solutions.
    huh equating the equations? Sorry I'm confused.
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    (Original post by junayd1998)
    huh equating the equations? Sorry I'm confused.
    y = 3x-7 and y= 2px^2 -6px + 4p.

    so 3x-7 = 2px^2-6px +4p

    You know have a quadratic equation, the next steps are standard.
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    (Original post by junayd1998)
    huh equating the equations? Sorry I'm confused.
    lets say if the equations had solutions (if they intersected) then you would rearrange for something like y in both equations and equate them. When you equate them you will get a quadratic. Now if a quadratic has real solutions b^2-4ac>0 and if it has no real solutions it will be less then 0. So when you basically equate them you need to show it has no real solutions
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    (Original post by Asad_2015)
    lets say if the equations had solutions (if they intersected) then you would rearrange for something like y in both equations and equate them. When you equate them you will get a quadratic. Now if a quadratic has real solutions b^2-4ac>0 and if it has no real solutions it will be less then 0. So when you basically equate them you need to show it has no real solutions
    so do we rearrange 3x-7=2px^2-6px+4p ?

    into a quadratic ?
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    (Original post by junayd1998)
    so do we rearrange 3x-7=2px^2-6px+4p ?

    into a quadratic ?
    When you have a quadratic equation you make it equal to 0.
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    (Original post by NotNotBatman)
    When you have a quadratic equation you make it equal to 0.
    yeah so, 2px^2-6px+4p+7-3x = 0 ?
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    (Original post by junayd1998)
    yeah so, 2px^2-6px+4p+7-3x = 0 ?
    Try grouping the x terms.
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    (Original post by NotNotBatman)
    Try grouping the x terms.
    There are no like terms I'm confused.
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    (Original post by junayd1998)
    There are no like terms I'm confused.
    It would be easier to have all the x terms together, so we can say 2px^2 + (-6p-3)x + 4p+7 = 0
    Writing it like this makes it easier to see what the a,b and c values are.
    Now for no points of intersection, there are no real solutions so b^2-4ac < 0
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    (Original post by NotNotBatman)
    It would be easier to have all the x terms together, so we can say 2px^2 + (-6p-3)x + 4p+7 = 0
    Writing it like this makes it easier to see what the a,b and c values are.
    Now for no points of intersection, there are no real solutions so b^2-4ac < 0
    ohhh okay yeah i get it thanks

    and then for part b we just solve the quadratic equation given in part a ?
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    (Original post by junayd1998)
    ohhh okay yeah i get it thanks

    and then for part b we just solve the quadratic equation given in part a ?
    The quadratic inequality, yes.
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    (Original post by NotNotBatman)
    The quadratic inequality, yes.
    So part B would be 1/2<x<18/4?
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    How did you access this years maths paper? (Just out of curiousity)
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    (Original post by junayd1998)
    So part B would be 1/2<x<18/4?
    Yes, 18/4 can be simplified as 9/2.
 
 
 
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