Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 35

Math Help - Writing Math Paper 1 on Friday - Help Please

  1. #16
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by DivideBy0 View Post
    Are you sure Area_{Circle+Square} is right?
    To be honest, no i don't. But i don't know what else to do.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by DivideBy0 View Post
    Are you sure Area_{Circle+Square} is right?
    No it is not right. See if you can do it for him.


    Hint: You make a cut. That length is the total perimeter (circumfurence) of the shape you are about to bend into. Since you know its perimeter you can express its area.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Of course! Thats where i went wrong with the square. I made the perimeter the area. I will try it again later.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Okay here are a few things i need help with please...

    1. The roots of the equation  x^2 - kx + k + 3 = 0 is equal. Calculate the roots of the equation.

    I keep getting that the roots of this equation doesnt exist...

    2. g(x) = 2x + 1 and p(x) = g(x + k) where k is a constant and x - 1 is a factor of p(x). Calculate k.

    I got k = -3

    3. We have a graph y = -2x + 6 and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line y = x.

    I was thinking it could be: x = -2y + 6 ?
    But i'm really not sure.

    4. The function f is defined by f(x) = x^{log 100 x}
    Determine x if f(x) = 1000

    I said that:
     10^3 = x^{2x}
     3 = 2x log x

    But i dont know what to do next.



    Thanks for any help guys.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    1. The roots of the equation x^2 -kx +k + 3 = 0 is equal. Calculate the roots of the equation.

    You mean the roots are equal. Then the equation is a perfect square, in the form (x-a)^2.

    So it can be done by "completing the square", or by the Qudratic Formula.
    In the x = {-b +,-sqrt[b^2 -4ac]} / 2a,
    if the [b^2 -4ac] is zero, then x = -b/(2a) only. Two of them (equal), because a quadratic always has two roots.

    x^2 -kx +k + 3 = 0 --------------(i)
    x^2 -kx +(k+3) = 0
    The [b^2 -4ac] is k^2 -4(1)(k+3).
    equate that to zero,
    k^2 -4k -12 = 0
    Factor that,
    (k-6)(k+2) = 0
    k = 6 or -2.

    When k = 6, in (i),
    x^2 -6x +6 +3 = 0
    x^2 -6x +9 = 0
    (x-3)^2 = 0
    x = 3

    When k = -2, in (i),
    x^2 +2x -2 +3 = 0
    x^2 +2x +1 = 0
    (x+1)^2 = 0
    x = -1

    Therefore, for the two roots to be equal, the equation (i) has 2 roots of 3, or 2 roots of -1. ----answer.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Thanks ticbol.
    I realised my mistake. I dont know why, but i accidentally wrote that Delta = b^2 + 4ac
    Follow Math Help Forum on Facebook and Google+

  7. #22
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    2.
    g(x) = 2x +1 and p(x) = g(x+k) where k is a constant and (x-1) is a factor of p(x). Calculate k.


    p(x) = 2(x+k) +1 ------(i)

    if (x-1) is a factor of p(x), then the other factor must have a degree of zero because p(x) is linear only (degree of 1) and (x-1) is of degree 1 already. So the other factor is a constant, say "a".

    a(x-1) = 2(x+k) +1
    ax -a = 2x +2k +1
    ax -a = 2x +(2k +1)
    For them to be equal, it must be that
    ax = 2x ---------------(1)
    -a = 2k +1 -----------(2)

    ax = 2x
    So, a = 2
    Substitute that into (2),
    -2 = 2k +1
    -2 -1 = 2k
    k = -3/2 ------------------------answer.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    3. We have a graph y = -2x +6 and we are asked to give the equation of the graph symmetrical to the above one, with respect to the line y = x.

    I was thinking it could be: x = -2y +6?
    But i'm really not sure.

    y = -2x +6
    y = -2(x-3)
    Swap x and y,
    x = -2(y -3)
    -x/2 = y -3
    -x/2 +3 = y
    or,
    y = -(1/2)x +3 ----------------the equation of that line.

    -------------------------------

    4. The function f is defined by f(x) = x^(log[100x])
    Determine x if f(x) = 1000.

    I said that: 10^3 = x^(2x)


    But log[100x] is not equal to 2x.
    It is equal to log[100] +log[x] = 2 + log[x].

    x^(log[100x]) = 1000
    Take the logs of both sides,
    (log[100x])*log[x] = log[10^3]
    (log[100] +log[x])*log[x] = 3
    (2 +log[x])*log[x] = 3
    2log[x] +(log[x])^2 = 3
    For simplicity, let y = log[x],
    2y +y^2 = 3
    y^2 +2y -3 = 0
    (y+3)(y-1) = 0
    y = -3 or 1
    Or,
    log[x] = -3 or 1

    log[x] = -3
    x = 10^(-3) = 1/1000 --------------answer.

    log[x] = 1
    x = 10^1 = 10 ----------------------answer also.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    We are given the following function:
    f(x) = 2x^3 + px^2 + qx + 3

    And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.

    ---

    1. Show that p = -5 and q = -4


    I set point F into f(x) and got: 2q + 4p = -28

    When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set p = -5 and q = -4 into 2q + 4p = -28 and see if the equation is true?
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Sorry this is so late lol

    Let L - x and x be the lengths for the circle and the square respectively.

    Then,
    2\pi r=L-x

    r=\frac{L-x}{2\pi}

    r^2=\frac{(L-x)^2}{4\pi^2}
    Area_{Circle}=\pi r^2=\frac{(L-x)^2}{4\pi}

    Also,
    Area_{Square}=\left( \frac{x}{4} \right)^2=\frac{x^2}{16}

    So
    Area_{Circle}+Area_{Square}=\frac{(\pi+4)x^2-8Lx+4L^2}{16\pi}

    \frac{dy}{dx}=\frac{(4+\pi)x-4L}{8\pi}

    Equating to zero and solving: x=\frac{4L}{4+\pi}
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Quote Originally Posted by janvdl View Post
    We are given the following function:
    f(x) = 2x^3 + px^2 + qx + 3

    And we are told that point F with co-ordinates of (2 ; -9) is a local minimum.

    ---

    1. Show that p = -5 and q = -4


    I set point F into f(x) and got: 2q + 4p = -28

    When they say show then it means something completely different than prove doesnt it? Does it mean i can just go ahead and set p = -5 and q = -4 into 2q + 4p = -28 and see if the equation is true?
    \frac{dy}{dx}=6x^2+2px+q

    Equating to zero and using the quadratic formula we get

    x=\frac{-p \pm \sqrt{p^2-6q}}{6}
    We have two solutions, but since we know that in a cubic graph when x^3 has a coefficient >0 the minimum is farther right than the maximum (hopefully someone with more experience can explain this better ). Hence, the x coordinate of the minimum is:
    \frac{-p+\sqrt{p^2-6q}}{6}

    So you have two simultaneous equations:

    \frac{-p+\sqrt{p^2-6q}}{6}=2

    2q+4p=-28

    Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by DivideBy0 View Post
    \frac{dy}{dx}=6x^2+2px+q

    Equating to zero and using the quadratic formula we get

    x=\frac{-p \pm \sqrt{p^2-6q}}{6}
    We have two solutions, but since we know that in a cubic graph when x^3 has a coefficient >0 the minimum is farther right than the maximum (hopefully someone with more experience can explain this better ). Hence, the x coordinate of the minimum is:
    \frac{-p+\sqrt{p^2-6q}}{6}

    So you have two simultaneous equations:

    \frac{-p+\sqrt{p^2-6q}}{6}=2

    2q+4p=-28

    Show assumes you don't know that p=-5 and q=-4 and you have to work them out from scratch.
    I thought of doing simulataneous equations, but i was a little lazy I've been doing maths since this morning


    Hey, i'll let you guys know about my math mark after the exam ok? Thanks for all the help.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by DivideBy0 View Post
    Sorry this is so late lol

    Let L - x and x be the lengths for the circle and the square respectively.

    Then,
    2\pi r=L-x

    r=\frac{L-x}{2\pi}

    r^2=\frac{(L-x)^2}{4\pi^2}
    Area_{Circle}=\pi r^2=\frac{(L-x)^2}{4\pi}

    Also,
    Area_{Square}=\left( \frac{x}{4} \right)^2=\frac{x^2}{16}

    So
    Area_{Circle}+Area_{Square}=\frac{(\pi+4)x^2-8Lx+4L^2}{16\pi}

    \frac{dy}{dx}=\frac{(4+\pi)x-4L}{8\pi}

    Equating to zero and solving: x=\frac{4L}{4+\pi}
    I sort of had the idea... OK, i was wrong! Good one DivideBy0!

    EDIT: Hey no, wait, this morning i started with it, except that my formulas were just the other way round.
    Like the square was (L - x), etc.
    But i stopped halfway through it because i had work to do...
    But i did it the way you did...
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Well tomorrow is D-Day for me...

    Anyway, I have one more question.

    For which values of k will k = 2x^3 + px^2 + qx + 3 have only one root? (There has been proven that p = -5 and q = -4)

    I've thought about using Delta, and equating it to zero, but that only applies to quadratic equations. This question only counts 2 marks, so the solution should be quite simple.
    Follow Math Help Forum on Facebook and Google+

  15. #30
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by janvdl View Post
    Well tomorrow is D-Day for me...

    Anyway, I have one more question.

    For which values of k will k = 2x^3 + px^2 + qx + 3 have only one root? (There has been proven that p = -5 and q = -4)
    And what are p\mbox{ and }q.

    I relized something interesting,
    Given x^3+ax^2+bx+c=0 which has exactly 1 root.
    It is necessary (thought maybe not sufficient) that b^3=a^3c.
    Can you prove that?
    It might help you.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. help with writing math expressions
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 4th 2009, 01:30 AM
  2. which software u use for math equation writing
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 22nd 2009, 12:45 AM
  3. 100cr digit number writing a paper
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: October 5th 2008, 02:49 AM
  4. Help with math paper
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 24th 2008, 01:59 PM
  5. [SOLVED] Math paper
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 12th 2005, 11:34 AM

Search Tags


/mathhelpforum @mathhelpforum