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    I am stuck on a disguised quadratic question - part iii) b), below.

    I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
    I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

    Thanks.

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    (Original post by alde123)
    I am stuck on a disguised quadratic question - part iii) b), below.

    I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
    I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

    Thanks.

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    k^\frac{1}{2} is the positive square root of k. So we have

    \sqrt{k} = -5

    k=25 is not a solution to this. Actually this equation has no solutions.
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    (Original post by Middriver)
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    Part a of the question says k>1
    That doesn't make a difference for this question.
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    (Original post by notnek)
    That doesn't make a difference for this question.
    Why is k^1/2 the positive root and not just the root of k?
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    (Original post by Middriver)
    Why is k^1/2 the positive root and not just the root of k?
    f(x)=x^\frac{1}{2} is equivalent to the function f(x)=\sqrt{x}.

    Both are single valued functions (as all functions are) that return the positive square root of the input.

    All positive numbers n have 2 square roots: n^\frac{1}{2} and -n^\frac{1}{2} where n^\frac{1}{2} is the positive square root of n.
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    Thanks notnek and Middriver, but I'm still a bit confused.
    Surely when you get k^1/2 = -5 and you square it, you get k = 25 which is > 1 so counts as a +ve solution.

    I've put a link to the mark scheme below
    http://www.ocr.org.uk/Images/61191-mark-scheme-june.pdf
    Its core 2 question 9 part iii) b)
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    (Original post by alde123)
    I am stuck on a disguised quadratic question - part iii) b), below.

    I managed to get to (y+5)(y-3) after substituting k^(1/2) with y so y=-5 or 3.
    I then got k=25 or 9 as the final answer but the mark-scheme only gives 9 as the final answer. Could someone please explain why.

    Thanks.

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    Wait what exam board are you doing... i thought all Core 2's are over? o.0
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    (Original post by Someboady)
    Wait what exam board are you doing... i thought all Core 2's are over? o.0
    I'm doing OCR but I'm in year 12 - my school's scrapped linear A-levels and is giving us end of year mocks instead.
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    (Original post by alde123)
    I'm doing OCR but I'm in year 12 - my school's scrapped linear A-levels and is giving us end of year mocks instead.
    I don't follow? Maths is still a modular subject as far as I know. You mean they're starting linear A-levels and they're making you do all of A-level Maths over 2 years?
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    (Original post by alde123)
    Thanks notnek and Middriver, but I'm still a bit confused.
    Surely when you get k^1/2 = -5 and you square it, you get k = 25 which is > 1 so counts as a +ve solution
    Whenever you a square both sides of an equation, the solutions you obtain are not guaranteed to be correct - you should always check them.

    Example : Solve x-1 = 2

    Square both sides :

    x^2-2x+1 = 4 \Rightarrow  x^2-2x-3 = 0 \Rightarrow (x+1)(x-3) = 0 \Rightarrow x = -1, x=3


    For your equation \displaystyle k^\frac{1}{2}=-5:

    k = 25 is not a solution since \displaystyle 25^\frac{1}{2} = 5.
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    (Original post by Someboady)
    I don't follow? Maths is still a modular subject as far as I know. You mean they're starting linear A-levels and they're making you do all of A-level Maths over 2 years?
    Yup.
    There is still an AS in the current specification right now, but (I don't really know how this works) my school decided that starting with my year group, everyone is going to sit the whole A-Level in Year 13. The only people who sat the their A-Level in maths this year, in my school, were the people doing Further Maths (which I'm not doing).
    Not all schools have gone linear though.
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    (Original post by alde123)
    Yup.
    There is still an AS in the current specification right now, but (I don't really know how this works) my school decided that starting with my year group, everyone is going to sit the whole A-Level in Year 13. The only people who sat the their A-Level in maths this year, in my school, were the people doing Further Maths (which I'm not doing).
    Not all schools have gone linear though.
    If you are still stuck after reading Notnek's explanations, have a look at this thread.

    What the graph shows is that there is no value k such that k^1/2 = -5.
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    (Original post by notnek)
    Whenever you a square both sides of an equation, the solutions you obtain are not guaranteed to be correct - you should always check them.

    Example : Solve x-1 = 2

    Square both sides :

    x^2-2x+1 = 4 \Rightarrow  x^2-2x-3 = 0 \Rightarrow (x+1)(x-3) = 0 \Rightarrow x = -1, x=3


    For your equation \displaystyle k^\frac{1}{2}=-5:

    k = 25 is not a solution since \displaystyle 25^\frac{1}{2} = 5.
    Thanks, that makes so much more sense now.
 
 
 
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