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Math Help - [SOLVED] Help Needed...

  1. #1
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    [SOLVED] Help Needed...

    -Show that x-a is a factor of x^n-a^n for any positive integer n.

    -Show that x+a is a factor of x^n+a^n for any positive odd integer n.



    Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.


    Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

    Consider the following system of equations in x+y:

    ax+by=c
    dx+ey=f

    Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?
    Last edited by CaptainBlack; June 9th 2009 at 02:16 PM. Reason: restore deleted post
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  2. #2
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    Quote Originally Posted by Szpieg View Post
    -Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.
    The slope of the line y = x is 1. The slope of the line with the points (a, b) and (b, a) is
    \frac{y_2 - y_1}{x_2 - x_1} = \frac{a - b}{b - a} = -1 ,
    so this line is perpendicular to the line y = x. (Slopes of perpendicular lines are negative reciprocals of each other.)

    Pick a point (c, c) that is on the line y = x. Use the distance formula to find the distance from (a, b) to (c, c), and from (b, a) to (c, c).


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  3. #3
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    thanks for the help, if anyone can help we with the others, it would be much appreciated...
    Last edited by CaptainBlack; June 9th 2009 at 02:17 PM. Reason: restore deleted post
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  4. #4
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    Quote Originally Posted by Szpieg View Post
    -Consider the following system of equations in x+y:

    ax+by=c
    dx+ey=f

    Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?
    If the constants a, b, c, d, e, f form an arithmetic sequence, then
    b = a + n
    c = a + 2n
    d = a + 3n
    e = a + 4n
    f = a + 5n
    for some n. Substitute:

    \begin{aligned}<br />
ax\ &+ &(a + n)y\ &=\ a + 2n \\<br />
(a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n<br />
\end{aligned}

    Solve by elimination. Multiply the 1st equation by -(a + 3n)/a:

    \begin{aligned}<br />
-(a + 3n)x\ &- &\frac{(a + n)(a + 3n)}{a}y\ &=\ -\frac{(a + 2n)(a + 3n)}{a} \\<br />
(a + 3n)x\ &+ &(a + 4n)y\ &=\ a + 5n<br />
\end{aligned}

    Add the 2 equations to cancel out the x's. The y coefficients, added together, give us

    - \frac{(a + n)(a + 3n)}{a} + (a + 4n) = - \frac{a^2 + 4an + 3n^2}{a} + \frac{a^2 + 4an}{a} = -\frac{3n^2}{a} ,

    and the constant terms, added together, give us

    - \frac{(a + 2n)(a + 3n)}{a} + (a + 5n) = - \frac{a^2 + 5an + 6n^2}{a} + \frac{a^2 + 5an}{a} = -\frac{6n^2}{a} .

    The two equations added together, therefore, is

    -\frac{3n^2}{a}y = -\frac{6n^2}{a} ,

    meaning that y = 2.

    Substitute y = 2 into either of the original equations (I chose the first):

    \begin{aligned}<br />
ax + (a + n)y &= a + 2n \\<br />
ax + (a + n)(2) &= a + 2n \\<br />
ax + 2a + 2n  &= a + 2n \\<br />
ax &= -a<br />
\end{aligned}

    So x = -1. We therefore have one solution to the system of equations, (-1, 2).


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    Last edited by yeongil; June 6th 2009 at 06:09 AM. Reason: Aligning equations in LaTeX
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  5. #5
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    OMG, thank you so much, this one i had no idea how to do=] ull be getting thanks from me =] and respect...by the way how old are you and what school do you go to?

    PS, i got the same answer, but not in a true proof...i used multiple examples, but my teacher does not want that...this is perfect
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  6. #6
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    Quote Originally Posted by Szpieg View Post
    -Show that x-a is a factor of x^n-a^n for any positive integer n.

    -Show that x+a is a factor of x^n+a^n for any positive odd integer n.



    -Prove that if a≠b, then (a, b) and (b, a) lie on a line perpendicular to the line y=x and are equidistant from y=x.


    -Prove that if f is a one-to-one odd function, f-^-1 is an odd function.

    -Consider the following system of equations in x+y:

    ax+by=c
    dx+ey=f

    Where the constants a, b, c, d, e, and f form an arithmetic sequence. How might you characterize the solutions to the system of equations?
    For your first question, call your function f(x).

    So f(x) = x^n - a^n.

    By the Factor and Remainder Theorems, (x - a) is a factor IF f(a) = 0.


    So f(a) = a^n - a^n = 0.

    So (x - a) is a factor of x^n - a^n.


    An alternative proof is done using factorisation.

    For any a, b

    a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1}).

    Applying this knowledge to f(x) we find

    f(x) = (x - a)(x^{n - 1} + x^{n - 2}a + x^{n -3}a^2 + \dots + x^2a^{n - 3} + xa^{n - 2} + a^{n - 1})

    So it can clearly be seen that (x - a) is a factor.




    For your second problem, to show (x + a) is a factor of x^n + a^n for any n that is an odd positive integer, do the same thing.


    Let f(x) = x^n + a^n.

    By the Factor and Remainder Theorems,
    (x + a) is a factor if f(-a) = 0.


    So f(-a) = (-a)^n + a^n.


    If n is even, you have f(-a) = a^n + a^n = 2a^n, so x - a is not a factor if n is even,


    BUT If n is ODD, you have f(-a) = -a^n + a^n = 0.


    So (x + a) is a factor of f(x) as long as n is a positive odd integer.



    Alternatively, use a similar alternate proof to the one above.



    If n is an odd integer, then


    a^n + b^n = (a + b)(a^{n - 1} - a^{n - 2}b + a^{n - 3}b^2 - \dots + a^2b^{n - 3} - ab^{n - 2} + b^{n - 1}).



    So f(x) = x^n + a^n = (x + a)(x^{n - 1} - x^{n - 2}a + x^{n - 3}a^2 - \dots + x^2a^{n - 3} - xa^{n - 2} + a^{n-1}).


    Here you can clearly see that x + a is a factor of f(x), but only if n is a positive odd integer.[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
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