Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$. How would one solve for x in the following equation:
$\dfrac{(1+x)^{36} -1}{x} =20142.9/420$
I tried factorising the top but that didnt really help much.
$((1+x)^{18} - 1)((1+x)^{18}+1)$
Any help is appreciated thanks.
 A: This is a standard example of an equation
that does not have an explicit formula for its root.
It can only be solved numerically.
The important thing is to get a good first approximation to the root.
For example, if you want to solve
$\frac{(1+x)^n-1}{x} = a$,
and you think that there might be a root close to $0$,
you can approximate $(1+x)^n$
by $1+nx+n(n-1)x^2/2$
(the first terms of the binomial theorem)
to get
$n + n(n-1)x/2 = a$
or $x = 2(a-n)/(n(n-1))$.
Then apply Newton's iteration.
A: Make a change in variables: $y = x + 1$. Then the equation becomes:
$$\dfrac{y^{36} - 1}{y-1} = 20142.9/420$$
The numerator $y^{36}-1$ is equal to $(y-1)(y^{35}+y^{34}+\cdots+1)$ (use the formula for the sum of a geometric series). Therefore dividing by $y-1$ gives:
$$y^{35}+y^{34}+\cdots+1 = 20142.9/420$$
This is a polynomial of degree 35 in $y$. It happens to be that for polynomials of degree $5$ or higher there is no general formula to solve for $y$ (1). You can find an approximate solution using a tool like Wolfram Alpha. Try inputting "(y^(36) - 1)/(y-1) = 20142.9/420" and see what answers you get for $y$ (there may be more than one). Then use $x = y - 1$ to get approximate answers for $x$.
A: Assuming we want a small positive value for $x$, we can approximate $(1+x)^{36}$ as $$1^{36}+{36 \choose 1} 1^{35}x^1 + {36\choose 2} 1^{34} x^2+O(x^3)$$
With this approximation, the equation becomes approximately $$\frac{36x+630x^2}{x}=47.959$$
This can be rearranged as $630x=47.959-36$, which has solution $x\approx 0.01898$.
