how can I calculate $r$ in $r + r^2 + r^3 + r^4 = 100$?

Question

As the title says, I want to know how to solve this equation. Just cannot find a way

I want to know how much % profit someone has to make EVERY MONTH in average from a stock within $4$ months, to double the money. In other words: I have to make r % profit on every single month of the 4 month to double (r=100%) the money.

Answer 1

The underlying problem is that you've got the wrong equation.

If you make a profit of $r$% after each month, and started with $n$, then after one month, you'd have $n\left(1+\frac{r}{100}\right)$, after two months, $n\left(1+\frac{r}{100}\right)^2$, and in general, after $m$ months:

$$n\left(1+\frac{r}{100}\right)^m$$

When $m=4,$ you want to have $2n$, so you want:

$$\left(1+\frac{r}{100}\right)^4=2$$

which yields:

$$r=100(\sqrt[4]2-1)\approx 18.92$$

Answer 2

Why would you need to know something like this: $$ -1/4-1/12\,\sqrt {3}\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3] {2260+6\,\sqrt {48382278}}}}}+1/12\,\sqrt {6}\sqrt {{\frac {-2\,\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [ 3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt {48382278}} }}} \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}+15\,\sqrt {3}\sqrt [3]{2260+6\,\sqrt {48382278}}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}} \sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\, \sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt { 48382278}}}}}+2404\,\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3] {2260+6\,\sqrt {48382278}}}}}}{\sqrt [3]{2260+6\,\sqrt {48382278}} \sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\, \sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt { 48382278}}}}}}}} $$

when numerical solution will give you $2.849217207$ approximately.

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Edit:

But in all seriousness you must factor and then use alternative methods such a Newton Raphson to solve for the zeroes. Otherwise its virtually impossible by just hand