Solve $(1+x)^2=A\sqrt{1+Cx}$ for $x$.

Question

$x>0$, $A>0$ and $C>1$.

I am trying to come up with a closed form expression for $x$, even if it is an approximation.

Any help appreciated.

Answer 1

Squaring your equation and expanding we get $$x^4+4x^3+6x^2+4x-A^2Cx+1-A^2=0$$ It is a polynomial of degree four.I think you will need a numerical method, if values for the parameters are given.

Answer 2

Square both sides and you get a quartic equation for $x$. Quartics do have closed-form solutions, but they are not pretty.

If you want an approximation, you should tell us what parameters can be considered as "small" or "large".

Answer 3

Numerically you can solve it using iterative process. Take initial approximation for your root, let's say $x_0=1$ and calculate the next approximation using this formula: $$x_k=\sqrt{A\sqrt{1+Cx_{k-1}}}-1$$ Stop the process when $|x_k-x_{k-1}|$ is small enough for your application.