I have the following equation:138000(1+x)5+71000(1+x)4+54000(1+x)3+37000(1+x)2+200001+x−20000=0

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Nunnaxf18 · Accepted Answer

This looks like an &quot;internal rate of return&quot; question, where usually you want x≻1When x≻1, the function is strictly decreasing as x increases. Clearly, if x is very large, then the value is negative, and when x=0 it is 20000, so you have one positive real root. You can use numerical methods to find the solution - binary search for example.I don&#039;t know what you&#039;d use on a mid-term - that would depend on what calculation tools you were allowed during the midterm.One quick way is to write t=11+x and note that we are solving g(t)=0 for some polynomial with g(1)=20000 and g′(1)=730000. So an estimate is t≈1−273 or x≈0.028. This is Newton&#039;s method, just on a slightly easier formula when done relative to t, and it converges quickly because the root x is close to 0. This might not work in general.Wolfram alpha gives the root as x≈0.02932

Roy Brady · Accepted Answer

Letting 1+x=t, multiplying with t5 and sorting, we findt5=20t4+37t3+54t2+71t+38200This suggests an iterative method by letting tn+1=20tn4+37tn3+54tn2+71tn+382005. A suitable starting value is t=1, leading to a solution t≈1.0923 and x≈−0.02849

I have the following equation: \(\displaystyle{\frac{{{138000}}}{{{\left({1}+{x}\right)}^{{5}}}}}+{\frac{{{71000}}}{{{\left({1}+{x}\right)}^{{4}}}}}+{\frac{{{54000}}}{{{\left({1}+{x}\right)}^{{3}}}}}+{\frac{{{37000}}}{{{\left({1}+{x}\right)}^{{2}}}}}+{\frac{{{20000}}}{{{1}+{x}}}}-{20000}={0}\)

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