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    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    4.B.III

    For this a subbed x in for -3 and hence that = 0 then it is a stationary point, mark scheme credits use of discriminant being negative therefore not having any/real roots.

    Since when was this the method? what does that mean? i must have missed this part of lesson or something.
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    (Original post by ayylmao12)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    4.B.III

    For this a subbed x in for -3 and hence that = 0 then it is a stationary point, mark scheme credits use of discriminant being negative therefore not having any/real roots.

    Since when was this the method? what does that mean? i must have missed this part of lesson or something.
    I think you have looked at the wrong mark scheme as that makes no sense
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    (Original post by ayylmao12)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    4.B.III

    For this a subbed x in for -3 and hence that = 0 then it is a stationary point, mark scheme credits use of discriminant being negative therefore not having any/real roots.

    Since when was this the method? what does that mean? i must have missed this part of lesson or something.
    Substituting x=-3 does not show that there are no other stationary points. You need to divide the cubic by (x + 3) and show that the resulting quadratic does not have any real roots.
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    (Original post by TenOfThem)
    I think you have looked at the wrong mark scheme as that makes no sense
    It makes sense to me.
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    (Original post by ayylmao12)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    4.B.III

    For this a subbed x in for -3 and hence that = 0 then it is a stationary point, mark scheme credits use of discriminant being negative therefore not having any/real roots.

    Since when was this the method? what does that mean? i must have missed this part of lesson or something.
    Notice the use of the word only in the question. You need to show that there are no other solutions of the cubic.

    You've already split up the cubic in 4ii so you need to show that the quadratic you've found has no solutions. That's where the discriminant becomes involved.
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    (Original post by Mr M)
    It makes sense to me.
    Durr

    I didn't think about the fact that he was referring to the quadratic once divided
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    Oh right its using the quad equation from the pevious equation, ah that makes sense now,
 
 
 
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