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Math Help - Geometric Series and the intersection of two graphs.

  1. #1
    Member rtblue's Avatar
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    Geometric Series and the intersection of two graphs.

    Questions:

    1. How many solutions (x,y) are there for the system of equations:

    \displaystyle x^2+y^2+3x-14y+30=0
    \displaystyle y^2-7xy+10x^2=0

    2. In a geometric series of positive terms, each term is the sum of the two following terms. What is the common ratio?

    Both of these questions are meant to be done without calculators.

    I have no clue where to start on number 1. On number 2, I tried setting up a system of equations, but fell one equation short:

    let a,b, and c be three consecutive terms of the series.

    a=b+c

    and

    c/b = b/a

    are the two equations I came up with, but they are not enough to solve the system.

    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by rtblue View Post
    2. In a geometric series of positive terms, each term is the sum of the two following terms. What is the common ratio?
    The common ratio behaves like this: r^n=r^{n+1}+r^{n+2}.
    Solve for r.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rtblue View Post
    1. How many solutions (x,y) are there for the system of equations:

    \displaystyle x^2+y^2+3x-14y+30=0
    \displaystyle y^2-7xy+10x^2=0
    You could graph them. Otherwise take a close look at the second equation.
    \displaystyle y^2-7xy+10x^2=0

    Notice that it factors:
    \displaystyle y^2-7xy+10x^2= (y - 5x)(y - 2x) = 0

    Thus we have two equations for y which are both lines: y = 5x and y = 2x. It turns out that both lines intersect the graph of the first equation in two places. (Prove this!) So you have four solutions.

    -Dan
    Last edited by topsquark; March 20th 2011 at 11:47 AM.
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  4. #4
    Member rtblue's Avatar
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    Thanks Dan! That really helped. Now, if I'm not mistaken, I'll substitute y = 5x and y = 2x into the first equation. Then once I solve from there I should have 4 solutions, as each substitution results in a quadratic.

    On a side note, I figured out how to do the second one. For anyone who is curious:

    let a, ar, and ar^2 be three consecutive terms in the series.

    a = ar + ar^2

    a = ar(1+r)

    1 = r+r^2

    r^2+r-1 = 0

    Solving the quadratic gives the ratio.

    Thanks again Dan!
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