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P1 maths - discriminant/top-up fees watch

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    i think you'll find this a fairly easy question (i say easy because its worth 3 marks) but i'm stuck so please can you explain every step

    "given that the quadratic equation

    2x^2 + 8x + (k+3) = 0

    has equal roots, find the value of the constant k"

    thanks
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    if the roots are equal, then b^2 - 4ac = 0
    8^2 - (4*2)(k + 3) = 0
    64 - 8k + 24 = 0
    88 = 8k
    k = 11
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    thanks for the help, do you know of any websites that are good for this type of maths (discriminant etc)?
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    (Original post by greenie787)
    thanks for the help, do you know of any websites that are good for this type of maths (discriminant etc)?
    nope, fraid not. dont you have a p1 textbook?
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    yeah but i'm finding it hard to understand, i get most of the stuff its just this and comparing coeficients
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    (Original post by edders)
    if the roots are equal, then b^2 - 4ac = 0
    8^2 - (4*2)(k + 3) = 0
    64 - 8k + 24 = 0
    88 = 8k
    k = 11


    This may be wrong. I think you should check it. I get k=5.
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    (Original post by greenie787)
    yes i just checked the mark scheme and it is 5. can you write up how you did it please?


    2x^2 + 8x + (k+3) = 0

    b^2 - 4ac = 0 (equal roots)
    8^2 - 4.2(k + 3) = 0
    64 - 8(k + 3) = 0
    64 - 8k - 24 = 0
    8k = 40
    k = 5.
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    thank you
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    (Original post by Ralfskini)
    2x^2 + 8x + (k+3) = 0

    b^2 - 4ac = 0 (equal roots)
    8^2 - 4.2(k + 3) = 0
    64 - 8(k + 3) = 0
    64 - 8k - 24 = 0
    8k = 40
    k = 5.
    Yep I can confirm that is correct, k=5
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    what about when the roots are >0 or < 0, is it possible to solve then?

    also there is a similar question worth 1 mark asking with a reason to "state the number of real roots of"
    x^3 -3x^2+6=0
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    (Original post by Ralfskini)
    2x^2 + 8x + (k+3) = 0

    b^2 - 4ac = 0 (equal roots)
    8^2 - 4.2(k + 3) = 0
    64 - 8(k + 3) = 0
    64 - 8k - 24 = 0
    8k = 40
    k = 5.
    yes very clever. it was only a small mistake with the signs.
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    (Original post by edders)
    yes very clever. it was only a small mistake with the signs.


    Forgiven.
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    (Original post by greenie787)
    what about when the roots are >0 or < 0, is it possible to solve then?

    also there is a similar question worth 1 mark asking with a reason to "state the number of real roots of"
    x^3 -3x^2+6=0
    >0 means two real different roots.
    <0 means no real roots. you can solve it using complex numbers though, but thats beyond alevel.
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    (Original post by edders)
    yes very clever. it was only a small mistake with the signs.
    Edders do Unis look favourably on you doing a gap year?
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    (Original post by 2776)
    Edders do Unis look favourably on you doing a gap year?
    i dont think they care either way, tbh
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    (Original post by edders)
    i dont think they care either way, tbh
    Hmm, ans you arn't bothered about the extra tutition fees that you will be paying?
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    (Original post by 2776)
    Hmm, ans you arn't bothered about the extra tutition fees that you will be paying?
    im not paying extra tuition fees. im going in 2004.
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    (Original post by edders)
    im not paying extra tuition fees. im going in 2004.
    Yes, and tution fees are in place by 3006 arn't they? So in 2006 you will be paying an extra year of tuition fee.
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    (Original post by 2776)
    Yes, and tution fees are in place by 3006 arn't they? So in 2006 you will be paying an extra year of tuition fee.
    for people starting 2006, i believe. :confused:
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    (Original post by edders)
    for people starting 2006, i believe. :confused:
    lol, yes sorry.

    But arn't teh fees start for 2006 for all people including those who study then as well? So you would pay an extra years of tutition fees if you havnt taken a gap year...
 
 
 

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