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    No idea how to do this question someone helpppp thanks :P

    It's June 2009 C3 OCR MEI, Question 7 ii)

    I've done i) A and B but not sure how to go about this.

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    What does (B) tell you about whether x^2+xy+y^2 can be < 0. Also, under what circumstances can it equal 0.

    Now apply to (A).
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    (Original post by DFranklin)
    What does (B) tell you about whether x^2+xy+y^2 can be < 0. Also, under what circumstances can it equal 0.

    Now apply to (A).
    So is (B) saying that it can not be less than 0?

    I get that the expansion of (B) is the same as the second pair of brackets in (A) but not sure how to link that??
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    (Original post by azizadil1998)
    So is (B) saying that it can not be less than 0?
    Yes.

    I get that the expansion of (B) is the same as the second pair of brackets in (A) but not sure how to link that??
    Well, if you have two expressions F, G and you know F > 0 and G > 0, then what can we say about FG?

    Note that although you've seen that we can't have G < 0, you'll need to rule out the case where G = 0.
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    (Original post by DFranklin)
    Yes.

    Well, if you have two expressions F, G and you know F > 0 and G > 0, then what can we say about FG?

    Note that although you've seen that we can't have G < 0, you'll need to rule out the case where G = 0.
    Ohhh yeah ok so the composite function has to be >0, And how do I rule that out?
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    (Original post by azizadil1998)
    Ohhh yeah ok so the composite function has to be >0, And how do I rule that out?
    You've been given that x &gt; y so what does that tell you about x-y?

    When can the other factor of the product x^2+xy+y^2 be 0? Is that possible given the condition above?
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    (Original post by atsruser)
    You've been given that x &gt; y so what does that tell you about x-y?

    When can the other factor of the product x^2+xy+y^2 be 0? Is that possible given the condition above?
    Ok so x - y will be >0 then...?

    And it can't be 0 then can it?
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    (Original post by azizadil1998)
    Ok so x - y will be >0 then...?

    And it can't be 0 then can it?
    You need to find the conditions under which x^2+xy+y^2 can equal 0. You then need to show this can't possibly happen if x > y.
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    (Original post by DFranklin)
    You need to find the conditions under which x^2+xy+y^2 can equal 0. You then need to show this can't possibly happen if x > y.
    Ohh ok then, gonna try that one min
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    (Original post by DFranklin)
    You need to find the conditions under which x^2+xy+y^2 can equal 0. You then need to show this can't possibly happen if x > y.
    Isn't this condition when x=0? or y=? I'm not sure if I know what you mean If x>y this can never be 0 e.g. if x = 2 and y=1, it = 7.
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    (Original post by azizadil1998)
    Isn't this condition when x=0? or y=? I'm not sure if I know what you mean If x>y this can never be 0 e.g. if x = 2 and y=1, it = 7.
    You know that x^3-y^3=(x-y)((x+\frac{y}{2})^2+\frac{3y^2}  {4})

    But to have 0 = x^3-y^3=(x-y)((x+\frac{y}{2})^2+\frac{3y^2}  {4}) requires that either:

    x-y=0

    or:

    (x+\frac{y}{2})^2+\frac{3y^2}{4} = 0

    The first possibility has been ruled out by the condition x &gt;y. You now need to rule out the second possibility. To start to do this note that if A \ge 0, B \ge 0, then to have A+B=0 requires that we have both A=0, B=0.

    Can you apply that fact to the expression in question?
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    (Original post by atsruser)
    You know that x^3-y^3=(x-y)((x+\frac{y}{2})^2+\frac{3y^2}  {4})

    But to have 0 = x^3-y^3=(x-y)((x+\frac{y}{2})^2+\frac{3y^2}  {4}) requires that either:

    x-y=0

    or:

    (x+\frac{y}{2})^2+\frac{3y^2}{4} = 0

    The first possibility has been ruled out by the condition x &gt;y. You now need to rule out the second possibility. To start to do this note that if A \ge 0, B \ge 0, then to have A+B=0 requires that we have both A=0, B=0.

    Can you apply that fact to the expression in question?
    Ok I get the first bit but I don't know how to apply this to the expression. I hateeee proof like literally do not get this at all! Didn't do this at AS since I did Edexcel now I have to practice it myself since my new school does OCR MEI and they occasionally put this type of dodgy question.
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    (Original post by azizadil1998)
    Ok I get the first bit but I don't know how to apply this to the expression. I hateeee proof like literally do not get this at all! Didn't do this at AS since I did Edexcel now I have to practice it myself since my new school does OCR MEI and they occasionally put this type of dodgy question.
    OK, let me spell it out a bit more:

    You want to rule out (x+\frac{y}{2})^2+\frac{3y^2}{4} = 0. So let's assume that it is true, and find conditions on x,y that make it true.

    First note that both terms are \ge 0 since they are squares. So they play the roles of A and B in what I posted earlier. That means that:

    (x+\frac{y}{2})^2+\frac{3y^2}{4} = 0 \Rightarrow (x+\frac{y}{2})^2 = 0, \frac{3y^2}{4} = 0

    Now solve those simultaneous equations.
 
 
 
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