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Maths A-Level Core 1 Simultaneous Equations (with discriminant) watch

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    Any help on the following question (7) would be greatly appreciated. Please see my working.
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    (Original post by OL350)
    Any help on the following question (7) would be greatly appreciated. Please see my working.
    You've made a slight mistake in your working.

    You got to:

    x^2 + kx^2 - 6kx + 4 = 0

    Instead of what you did next, you should have done:

    (1+k)x^2 - 6kx+4=0 and gone on from there. Then you can get the range of values for k.

    Can you see why?
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    Yes that totally makes sense. Collect the like terms is what you're saying? Going by that is the final part of my answer correct as I seem to have a different one to the book...?
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    (Original post by OL350)
    Going by that is the final part of my answer correct as I seem to have a different one to the book...?
    I'm afraid not.

    (-6k)^2 = 36k^2, not 36k.
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    Sorry, obviously it's 36k^2.

    That at leaves me with...

    36k^2 - 16k - 16 > 0
    I divided by 16
    2k^2 - k - 1 > 0
    Factorised
    (2k+1)(k-1)
    Yet still got a different answer to the book - (k < - 2/3(minus two over 3) square root of five) OR (k > 2/3 (positive two over three) square root of five).

    Again any help would be greatly appreciated (this is obviously something I'm struggling on). Was factorising the correct thing to do?
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    (Original post by OL350)
    Sorry, obviously it's 36k^2.

    That at leaves me with...

    36k^2 - 16k - 16 > 0
    I divided by 16
    2k^2 - k - 1 > 0
    Factorised
    (2k+1)(k-1)
    Yet still got a different answer to the book - (k < - 2/3(minus two over 3) square root of five) OR (k > 2/3 (positive two over three) square root of five).

    Again any help would be greatly appreciated (this is obviously something I'm struggling on). Was factorising the correct thing to do?
    36/16
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    Could you explain your answer please.
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    (Original post by OL350)
    Sorry, obviously it's 36k^2.

    That at leaves me with...

    36k^2 - 16k - 16 > 0
    I divided by 16
    2k^2 - k - 1 > 0
    Factorised
    (2k+1)(k-1)
    Yet still got a different answer to the book - (k < - 2/3(minus two over 3) square root of five) OR (k > 2/3 (positive two over three) square root of five).

    Again any help would be greatly appreciated (this is obviously something I'm struggling on). Was factorising the correct thing to do?
    Ah I've just spotted another mistake in your working that I didn't originally - you've squared (3 - kx) slightly incorrectly before plugging that in for y. See if you can find out where the mistake is and correct it.

    (Also, the answer in the book is slightly wrong.).
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    (Original post by OL350)
    Could you explain your answer please.
    (Original post by lyricalvibe)
    36/16
    I was just saying you divided wrong when you divided by 16 as 36/16=!2 but usy seems to be actually be helping you with the question so just listen to him


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    My bad, 36/2 obviously isn't 2.

    (3-Kx) (3-Kx) = 3x3 = 9 +
    3 x -Kx = -3Kx +
    3 x -Kx = -3Kx +
    -Kx x -Kx = +Kx^2

    I don't see anything Wrong?
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    (Original post by OL350)
    My bad, 36/2 obviously isn't 2.

    (3-Kx) (3-Kx) = 3x3 = 9 +
    3 x -Kx = -3Kx +
    3 x -Kx = -3Kx +
    -Kx x -Kx = +Kx^2

    I don't see anything Wrong?
    kx(kx) = k^2x^2 - don't forget that the k is squared too.
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    Ok, gathering all the information given I have come to a final answer (whether it is correct or not I'm unsure as you say the answer in the book is slightly wrong). Please let me know what you think and correct me where I'm wrong, I really need a solution so I can work on other similar questions.
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    (Original post by OL350)
    Ok, gathering all the information given I have come to a final answer (whether it is correct or not I'm unsure as you say the answer in the book is slightly wrong). Please let me know what you think and correct me where I'm wrong, I really need a solution so I can work on other similar questions.
    Nope. :no:

    20k^2 > 16

    When you take the square root, you need to take the root of both sides (i.e. the 20 should be square rooted too).

    Instead, it's be easier if you divided through by 20 so that you get:

    k^2 > 4/5


    So k &lt; -\dfrac{2\sqrt{5}}{5}, k &gt; \dfrac{2\sqrt{5}}{5}
 
 
 

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