Help with this question please :)

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    Any help with question 4 would be much appreciated! How do you find the points of intersection?

    Thanks
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    I wish u could say this to maths:

    animeamanda1412 someone needs ur help
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    (Original post by tamil fever)
    I wish u could say this to maths:

    animeamanda1412 someone needs ur help
    lmaoo, i can never say that
    im always gonna be like "maths, forever saty a cute little baby so i can always solve your questions"

    and yh, what do you need help with?
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    (Original post by JakeAntonyBrown)
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    Any help with question 4 would be much appreciated! How do you find the points of intersection?

    Thanks
    4a - solve for x (factorise) check it can be factorised with discriminant
    4b - then put them into the equation of a straight line to find m or something like that
    C1
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    (Original post by animeamanda1412)
    lmaoo, i can never say that
    im always gonna be like "maths, forever saty a cute little baby so i can always solve your questions"

    and yh, what do you need help with?
    It's question 4, how do you find the points of intersection?

    Thanks
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    (Original post by tamil fever)
    I wish u could say this to maths:

    animeamanda1412 someone needs ur help

    That's brilliant! I'd totally be saying that 90% of the time.
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    (Original post by CheeseIsVeg)
    4a - solve for x (factorise) check it can be factorised with discriminant
    4b - then put them into the equation of a straight line to find m or something like that
    C1
    Thanks! For 4a, how would I solve for x?

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    (Original post by JakeAntonyBrown)
    Thanks! For 4a, how would I solve for x?


    you solve for by letting y=0
    as x is common factor in all the three terms, you factorise it out and you will end up with a quadratic equation which I think you can factorise.
    hope this helps
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    (Original post by JakeAntonyBrown)
    Thanks! For 4a, how would I solve for x?

    You factorise.
    Start by y=x^3-4x^2+3x
    now what is common to all terms/what can you put outside the brackets?
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    (Original post by JakeAntonyBrown)
    That's brilliant! I'd totally be saying that 90% of the time.
    i say that 100% of the time.I've even written it in my tests
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    (Original post by CheeseIsVeg)
    You factorise.
    Start by y=x^3-4x^2+3x
    now what is common to all terms/what can you put outside the brackets?
    X is common in all?

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    (Original post by JakeAntonyBrown)
    X is common in all?

    yup so that gives:
    y=x(x^2-4x+3)
    Now this means one root x = 0 (when y= 0 ) Do you see how I got that?You have a quadratic to factorise now, I was hoping you could do that on your own to get the other two roots?
    (x^2-4x+3)
    Is this helping - I don't want to feed you answers without explanations if you see what I mean?
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    (Original post by CheeseIsVeg)
    yup so that gives:
    y=x(x^2-4x+3)
    Now this means one root x = 0 (when y= 0 ) Do you see how I got that?You have a quadratic to factorise now, I was hoping you could do that on your own to get the other two roots?
    (x^2-4x+3)
    Is this helping - I don't want to feed you answers without explanations if you see what I mean?
    I understand how we get to x(xˆ2-4x+3) but from there I'm lost.

    Sorry
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    (Original post by JakeAntonyBrown)
    I understand how we get to x(xˆ2-4x+3) but from there I'm lost.

    Sorry
    So y= x(xˆ2-4x+3) yep?Now if we seperate them into two parts, where the x intercepts are, y = 0
    Therefore 0=x(xˆ2-4x+3)
    Here you can find out that either x=0 (from the coefficient infront of the bracket)
    or 0=(xˆ2-4x+3)

    So you must factorise xˆ2-4x+3
    Can you factorise or do you want to learn the Cheese way :rofl:
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    tbh, these questions are my weak points in maths
    but i'll give a method im not sure you'd like
    my teacher taught me a quick way, but i could never grasp it, so i always draw the curve, its better
    so draw the curve on a piece of paper and from there you can get values of A, B and C
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    (Original post by CheeseIsVeg)
    So y= x(xˆ2-4x+3) yep?Now if we seperate them into two parts, where the x intercepts are, y = 0
    Therefore 0=x(xˆ2-4x+3)
    Here you can find out that either x=0 (from the coefficient infront of the bracket)
    or 0=(xˆ2-4x+3)

    So you must factorise xˆ2-4x+3
    Can you factorise or do you want to learn the Cheese way :rofl:
    (x-3) (x-1)

    I don't know where all my GCSE knowledge has gone, how did I manage to get an A* at GCSE and not know anything now?!? :P
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    (Original post by JakeAntonyBrown)
    (x-3) (x-1)

    I don't know where all my GCSE knowledge has gone, how did I manage to get an A* at GCSE and not know anything now?!? :P
    yep you got it, so three roots are?
    Are you good now? I've got to go and tackle C3 myself
    Good on you :yy: Just do a bit of revision, should be fine
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    (Original post by CheeseIsVeg)
    yep you got it, so three roots are?
    Are you good now? I've got to go and tackle C3 myself
    Good on you :yy: Just do a bit of revision, should be fine
    +1
    +3
    and emmm???

    Do you know what this topic would be called?

    Thanks
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    (Original post by JakeAntonyBrown)
    +1
    +3
    and emmm???

    Do you know what this topic would be called?

    Thanks
    Factorising/solving a quadratic equation
    and x=0
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    (Original post by CheeseIsVeg)
    Factorising/solving a quadratic equation
    and x=0
    Thanks, how did you know x=0?

 
 
 
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