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Math Help - 1 equation 1 unknown but still killing my brain

  1. #1
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    1 equation 1 unknown but still killing my brain

    hi all. i have the following equation and am not sure how to solve it. it seems so simple but i haven't done math in a while and am not sure the steps to solve this. I can create a number close enough using excel to solve it through trial and error which is enough to tell me that it is possible, but now i just want to know the steps/process:

    100+100x+100x^2+100x^3+100x^4 = 400

    Essentially I have the first month's 'investment', and the total at the end of 5 months. Since I know that each subsequent month's investment is smaller than the first, I am trying to solve for the linear rate of decline...please help, and let me know if you have questions on what i'm trying to achieve. The numbers above (100 and 400) are arbitrary and I just picked them entirely at random for the purpose of simplification.
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  2. #2
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    make the equation equal to 0 first, we get 100x^4+100x^3+100x^2+100x-300=0
    find a value for x for which the equation is true. that is find a value for x for which the RHS is 0. say x_1
    then (x-x_1) is a factor of the RHS. divide by this factor and you will get an expression to the third power for x. repeat above steps, and get another expression to the second power for x, then you can factorize it all.
    if the equation has all real roots, you will get something like this:
    (x-a)(x-b)(x-c)(x-d)=0
    a,b,c,d, then are the roots of the equation
    I hope i am clear enough
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  3. #3
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    Quote Originally Posted by shortround View Post
    hi all. i have the following equation and am not sure how to solve it. it seems so simple but i haven't done math in a while and am not sure the steps to solve this. I can create a number close enough using excel to solve it through trial and error which is enough to tell me that it is possible, but now i just want to know the steps/process:

    100+100x+100x^2+100x^3+100x^4 = 400

    Essentially I have the first month's 'investment', and the total at the end of 5 months. Since I know that each subsequent month's investment is smaller than the first, I am trying to solve for the linear rate of decline...please help, and let me know if you have questions on what i'm trying to achieve. The numbers above (100 and 400) are arbitrary and I just picked them entirely at random for the purpose of simplification.
    x^5-1=100*(x-1) - Wolfram|Alpha
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  4. #4
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    hi. you lost me at "say then is a factor of the RHS. divide by this factor"...

    i don't know what that means can you clarify? thanks a lot for your help
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  5. #5
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    thanks, but i'm actually more interested in the procedure more than the answer itself...
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by shortround View Post
    thanks, but i'm actually more interested in the procedure more than the answer itself...
    Did you look at the exact roots? You want an exact answer? Have fun.
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  7. #7
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    Hello, shortround!

    Are you sure of the wording of the problem?


    100+100x+100x^2+100x^3+100x^4 \:=\: 400

    Essentially I have the first month's investment, and the total at the end of 5 months.
    Since I know that each subsequent month's investment is smaller than the first,
    I am trying to solve for the linear rate of decline.

    If the rate of decline is linear (not geometric), you have the wrong equation.

    . . . \begin{array}{c|c}<br />
\text{Month} & \text{Deposit} \\ \hline<br />
1 & 100 \\<br />
2 & 100-d \\<br />
3 & 100-2d \\<br />
4 & 100-3d \\<br />
5 & 100-4d \\ \hline<br />
\text{Total} &  400 \end{array}


    We have: . 500 - 10d \:=\:400 \quad\Rightarrow\quad d \:=\:10


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  8. #8
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    Quote Originally Posted by Soroban View Post
    Hello, shortround!

    Are you sure of the wording of the problem?



    If the rate of decline is linear (not geometric), you have the wrong equation.

    . . . \begin{array}{c|c}<br />
\text{Month} & \text{Deposit} \\ \hline<br />
1 & 100 \\<br />
2 & 100-d \\<br />
3 & 100-2d \\<br />
4 & 100-3d \\<br />
5 & 100-4d \\ \hline<br />
\text{Total} &  400 \end{array}


    We have: . 500 - 10d \:=\:400 \quad\Rightarrow\quad d \:=\:10


    well the actual payments are multiplied by 1 single rate that is less than 100%, so instead of 100x +100x^2, it's technically

    Payment 1 + Payment 2 + Payment 3, where
    Payment 2= Payment 1x
    Payment 3= Payment 2x

    Since the rate never changes, wouldn't that be a linear decline?
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