# How do you solve for x in this equation?

I tried expanding, but I still can't get rid of the exponents to isolate x.

$$\frac{(1+x)^4-1}{x}=4.374616$$

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This becomes a cubic equation. Are you interested in the exact answer or a numerical approximation? – JimmyK4542 Aug 21 '14 at 4:28
With that terrible number on the right you're looking at numerical methods, probably. Newton's Method is nice and straightforward, but you need a little calculus to prime the pump. – Dan Uznanski Aug 21 '14 at 4:31
Actually, the exact answer is $x = 0.06$. See WolframAlpha – JimmyK4542 Aug 21 '14 at 4:31
An exact answer would be great. That's an amazing site by the way, but still doesn't show the steps how it arrived at x. I hope there's a way to find it without the need to plot points on a graph. – user170492 Aug 21 '14 at 4:55
There's a section on the Wikipedia article Cubic function that discusses different methods for solving cubic equations. – HelloGoodbye Aug 21 '14 at 12:19

You want to solve $$\frac{(1+x)^4-1}{x}=a$$ After expansion and simplification, this equation write $$x^3+4x^2+6x=a-4$$ you could solve using Cardano's formula. On the other side, you can consider that the solution is the intersection of the function $$y=x^3+4x^2+6x$$ with the horizontal line $y=a-4$. If you study the function, you can prove that it does not show any maximum since its derivative $$y'=3x^2+8x+6$$ has no real root. So, there is only a real solution which will be positive if $a \gt 4$, $0$ if $x=4$ and negative if $a \lt 4$.

Since in your case, $a$ is not much larger than $4$, the solution is close to $0$ and a Newton procedure can be used. Starting with an initial guess $x_0=0$, Newton will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Using $f(x)=x^3+4x^2+6x-0.374616$, the successive iterates will be $0.062436$, $0.0600038$, $0.0600000$ which is the exact solution.

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And observe that numeric methods are "unfairly" good at solving problems where the exact solution is a nice round number, because once you've got close you can confirm whether the nice round number it is indeed exact. You have no "right" to expect the iteration to produce the exact solution since a priori it might not be rational ;-) – Steve Jessop Aug 21 '14 at 9:07
If you use Cardano, you immediately find that the only real solution is $x=\frac{3}{50}$. Change the rhs to $4.374617$; its is much more funny; the solution is $0.0600001540642...$. – Claude Leibovici Aug 21 '14 at 9:10
Sure, I'm remarking on your alternative. – Steve Jessop Aug 21 '14 at 9:11
If you're still going to end up solving the equation numerically, why didn't you use Newton–Raphson from the beginning? – HelloGoodbye Aug 21 '14 at 12:12
@HelloGoodbye. I thought that it could be good to show first that there is only one real solution and that $0$ is a quite good approximation. – Claude Leibovici Aug 21 '14 at 12:16

\begin{align} (1+x)^{4} - 1 &= [(1+x)^{2} - 1] [ (1+x)^{2} + 1] = (1+x-1)(1+x+1)(x^{2} + 2x +1) \nonumber\\ &= x(x+2)(x^{2}+2x+2) \end{align} or \begin{align} \frac{(1+x)^{4} -1}{x} = (x+2) (x^{2} + 2x +2). \end{align} Now, $4.374616$ can be factored into \begin{align} 4.374616 = (2.06)(2.1236) = (2 + .06)((.06)^{2} + 2(.06) + 2). \end{align} This yields \begin{align} (x+2) (x^{2} + 2x +2) = (2 + .06)((.06)^{2} + 2(.06) + 2) \end{align} which gives $x = .06$.

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I think it's too specific if one is expected to factor a number like that. – justhalf Aug 21 '14 at 7:07
+1 for a nice solution – Silviu Burcea Aug 21 '14 at 12:36

I will present two general approaches, starting with the one that works best for this particular case:

Assuming small values of x, we have $(1+x)^4\approx1+4x+6x^2$. Subtracting $1$ and dividing by x, we are left with solving $4+6x=4+\alpha\iff x=\dfrac\alpha4=0.06,~$ where $\alpha=0.374616$.

Let $~f(x)=(1+x)^4.\quad$ Then $~\dfrac{(1+x)^4-1}x=\dfrac{(1+x)^4-(1+0)^4}{x-0}\approx f'(0).\quad$ But $~f'(x)=$

$=4\cdot(1+x)^3.\quad$ So $~4\cdot(1+x)^3\approx a\iff x\approx-1+\sqrt[3]{\dfrac a4}=0.03,~$ which unfortunately is

only half of the true value, $x=0.06.~$ Nevertheless, the reason I chose to present this method as well is because, despite not working very well for this particular case, it works quite well in many other similar situations.

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