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  1. #1
    Quiraikotsu is offline Senior Member Community Builder
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    Default HELP! (with maths). Please.

    I'm panicking.
    And I remembered there's maths geniuses on ST.

    The quadratic equation x^2 + 4kx + 3k = 0 has two distinctive roots. Find the set of possible values of the constant k.

    ----

    The equation x^2 4kx + 3k = 0, where k is a constant, has distinct real roots.
    a. Prove that k(4k - 3) > 0.

    b. Hence find the set of possible values of k.

    It is given instead that the x-axis is a tangent to the graph of y = x^2 _ 4kx + 3k.

    c. Write down the possible values of k.

    (I figured this all has something to do with that b^2 - 4ac thing. =___='

    Help?

    p.s. Can anyone explain how the algebra for even and odd functions of graphs work?
    I mean- the even graphs are:
    f(x) = -f(x)

    and odd graphs are:

    f(x) = -f(-x)

    thing.

    PLEASE.
    HELP.
    ME.
    I'm begging.

  2. #2
    irecinius is offline Senior Member Community Builder
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    Lemme do it on paper and scan it for you.. be back in 10min.
    "Chile is a thin and tall country"

  3. #3
    neruke is offline Senior Member Long Time Member
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    Default

    O.o;

    I'm an idiot in math... really... but even I can solve this... >.>
    We did stuff like this last year, or the year before that.

  4. #4
    Urameshi-sama is offline Senior Member Community Builder
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    If you use calculus, this is uber easy, but you can get this using algebra too. I'll use algebra since it seems to be that level of problem.

    1)remember that for quadratics of form Ax^2+Bx+C=0, there is the eqn x=-b/(2a). Plug that into this quadratic to get x=-4k/2=-2k.
    2) Plug in this value for all x's in the original quadratic to get 4k^2-8k^2+3k=-4k^2+3k=0. Rearrange the eqn to get k(4k-3)=0. k=0, 3/4. (ans to b and c)
    3) part A doesn't make sense because it isn't true...

    4) for the odd, even function thing, that statement refers to the x you plug in. In even eqns, you will get the same y or f(x) if you plug in a number and its negative number (such as 2 and -2). For odd functions, its the exact opposite, so if you plug in a number and its negative, one of the y's will be positive and the other one negative.

    Edit: I didn't see your post Ire...

  5. #5
    Kolox is offline Senior Member Community Builder
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    It extremely easy math problem, but why would you ask it on this particular forum, Quira ? I don't see much sense...

  6. #6
    Quiraikotsu is offline Senior Member Community Builder
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    Default

    I live in England.
    Of course I'm dumb at maths. See the logic?

    And I don't know any maths forums... and people helped me here with maths last year, so I thought I'd ask.

    Thanks for the help, though

  7. #7
    Compjotr is offline Senior Member Long Time Member
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    Quira, read here about even and odd functions.
    It's hard to explain math in english for me, so I googled ... :P

  8. #8
    kayangelus is offline Senior Member Always Around
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    x^2 4kx + 3k = 0

    x = (-4k +SQRT(16k^2 - 12k))/2 _________ Quadratic formula
    x = (-4k +2SQRT(4k^2 - 3k))/2 __________ Removed the 4 from within the square root side and moved it outside
    x = -2k +SQRT(4k^2 - 3k) ______________ divide through by 2
    x = -2k +SQRT(k(4k - 3)) ______________ rearranged the stuff within the square root sign
    k(4k - 3) > 0 _______________________________ if it is less than zero, you have imaginary roots. If k(4k - 3) is equal to zero, you have 1 distinct root.

    I'm not sure how you guys learned it, but for my class, we just make a # line at this point, and plot values of k that make the equation 0, and draw lines there, creating section. In each section, we see whether it is positive or negative.

    You find that k(4k-3) > 0 if k<0, or if k>3/4.

    so, for values of k, (-infinity , 0)u(3/4 , infinity)

    b. doesn't make sense, as that would mean there being only 1 distinct root.


  9. #9
    cpr's Avatar
    cpr
    cpr is offline Super Moderator Community Builder
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    Sorry, it took a while for me to host em:

    C is the only one i'm not sure if it's right or not.

    Granted, these might not be right or not. I've officially retired from math as of May 2007. (All my Math requirements are fulfilled.) ^_^

    Your homework question:


    edit: ignore me circling what "x" equals. I forgot the question didn't care what the x-values were.

    Odd and even functions:


    weeeeeeeeeeee! you get to see my uglyz handwriting! Sorry, I absolutely hate hate hate hate hate reading/typing math on the computer. It's harder to follow.

  10. #10
    irecinius is offline Senior Member Community Builder
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    Been 7 years that I used algebra.. I got so rusty, not to mention I couldn't get my scanner to work..


    Lets begin.

    D = b&#178; - 4a.c
    x = -b + Root[D] / 2.a

    x' = -4k + root[4(4k&#178; -3k)] /2
    x' = -4k + 2.root[4k&#178; -3k)] /2

    x' = -2k + root[4k&#178; -3k]
    x'' = -2k - root[4k&#178; -3k]

    Now

    k(4k -3) > 0
    k > 3/4

    Meaning for that to be true k have to be > then 3/4

    And f(x) is just that, one is positive, the other is negative, being the same opposite.

    take f(x) = ax&#178;, a =3
    you will have


    Well I just read the posts that people made in between of me trying to get my scanner to work and not getting it, and almost breaking it in half =p
    Its just a bunch of stupid formulas to remember, no secret.

    And just go by CPR, its alot better explained.. I have the bad habit of doing stuff in my head.
    Last edited by irecinius; 10-28-2007 at 11:56 AM.
    "Chile is a thin and tall country"

 

 
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