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Math Help - find x when y = 0

  1. #1
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    find x when y = 0

    Hi. I have a problem which I think should be pretty easy to solve, but I'm not sure how to go about solving it. If, when x = 1990, y = 25,165,824 when x = 2000, y = 393,216 and when x = 2010, y = 12,288 what is the value of x when y = 0? I want to say this is an example of exponential decay, but I haven't solved a problem like this in quite a while, and I'm just not sure. Thanks for the help,
    Risingstar64
    Also, if this is the wrong section or title, please feel free to change it, or inform me and I will change it (if I can).
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  2. #2
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    I don't think this is an exponential decay.
    First, x is incremented by same amount, 10 years, each step. However, y is not reduced by the same factor.
    25165824 / 393216 = 64 but
    393216 / 12288 = 32
    showing that y is not exponential.

    Also, if it was some kind of multiplicative reduction, y would never reach 0.

    Is there any more information for the problem?
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  3. #3
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    Unfortunately, no. The only other information I can provide is that think you are supposed to use excel and a trendline to solve the equation. I appologize, but I have to go somewhere for a few hours, so I won't be able to respond again until around 3:00 pm.
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  4. #4
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    If all you are given is three values of (x, y), there is no way to determine x in (x, 0). There are an infinite number of functions, even polynomials that go through three given points- any you can choose one that has any value of x you want for y= 0.

    The simplest thing to do is to find the unique quadratic polynomial that goes through those given points.

    If y= ax^2+ bx+ c then (1990, 25,165,824) gives 25164824= a(1990)^2+ b(1990)+ c when (2000, 393,216) gives 393216= a(2000)^2+ b(2000)+ c and (2010, 12,288) gives 12288= a(2010)^2+ b(2010)+ c. solve those three equations for a, b, and c and solve the resulting equation ax^2+ bx+ c= 0.
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  5. #5
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    Hi. I just got back, and that seems like a fine solution. I'm not asking for the answer, but how do you solve a quadratic equation for a, b, and c? I only know how to solve for x.
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  6. #6
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    Notice the 3 equations written by HallsOfIvy with only a, b, and c.

    How do we solve 3 equations with 3 unknowns? Forget about x being quadratic.
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  7. #7
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    I'm sorry, but I don't know how to solve 3 equations with 3 unknowns (at least, not when they are that huge). Anyway, I have to go for tonight, but I'll try to figure it out on my own from here. thanks everyone for all the help.
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  8. #8
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    Quote Originally Posted by risingstar64 View Post
    Hi. I just got back, and that seems like a fine solution. I'm not asking for the answer, but how do you solve a quadratic equation for a, b, and c? I only know how to solve for x.
    You don't. You solve three linear equations for a, b, and c.

    Notice that all of them have just "+ c". Subtracting one of the equations from another eliminates c. Subtracting a different pair also eliminates c and leaves two equations in two unknowns. Solve one equation for either a or b and put it into the other to get a single equation in a single unknown.
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  9. #9
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    \displaystyle <br />
\begin{bmatrix}<br />
1990^2 & 1990 & 1 & 25164824\\ <br />
2000^2 & 2000 & 1 & 393216\\ <br />
2010^2 & 2010 & 1 & 12288<br />
\end{bmatrix}

    Use elementary row operations--adding/subtracting two rows, multiplying a row by scalar, and multiplying by scalar and adding or subtracting--until you reach the reduced row echelon form.

    Spoiler:

    \displaystyle <br />
\begin{bmatrix}<br />
1 & 0 & 0 & \frac{609767}{5}\\ <br />
0 & 1 & 0 & \frac{-2445356134}{5}\\ <br />
0 & 0 & 1 & 490329246816<br />
\end{bmatrix}
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  10. #10
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    If you don't like those numbers, shift everything by 2000- that is, write your function as a(x- 2000)^2+ b(x- 2000)+ c. Tnen you have a(-10)^2+ b(-10)+ c= 25165824, c= 393216, and a(10)^2+ b(10)+ c= 12288. You know c from the middle equation so just subtract it from each side of the other two equations. Then one has "10b" while the other has "-10b". Just add the two equations to get a single equation in a. Since (-10)^2= 10^2 you can also subtract the two equations to get a single equation for b.

    Of course, you can use an exponential if you want- you just can't use " y= ae^{bx} because then y is never 0. You could use, instead, y= ae^{bx}+ c or, better, y= ae^{b(x- 2000)}+ c. Then your equations become
    ae^{-10b}+ c= 25165824
    ae^{0}+ c= a+ c= 393216
    and ae^{10b}+ c= 12288

    You can write those as
    ae^{-10b}= 25165824- c
    ae^{0}= a= 393216- c
    and ae^{10b}= 12288- c

    Then, say, divide the first equation by the second to get e^{-10b}= \frac{25165824- c}{393215- c} and divide the third by the second to get e^{10}b= \frac{12288- c}{393215- c}.
    Since e^{-10b}= \frac{1}{e^{10b}} that is the same as
    \frac{25165824- c}{393215- c}= \frac{393215- c}{12288- c}
    which reduces to a quadratic equation for c.
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