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.

Have you tried squaring both sides? I know it may add more solutions but doing something generally generates more interest in your question — Mandelbrot, Feb 27 '19 at 19:48
I've tried to solve the quartic equation and it is doable, if we can solve the following: $$ z (z + 1)(z +2) = AC^2 \ / \ (C-1)^{3/2} $$ At least one solution $z$ is needed, although getting all solutions makes later calculations easier. In further calculations $$ z\cdot A{(C-1)^{1/2}} $$ is used. Would this be any help? $\ $ I used the algorithm from [http://www.sosmath.com/algebra/factor/fac12/fac12.html ], see also [http://mathforum.org/dr.math/faq/faq.cubic.equations.html ]. — eudes, Mar 02 '19 at 08:06
@eudes or $$ w^3 - w = AC^2 \ / \ (C-1)^{3/2}, \quad w = z+1 $$ — eudes, Mar 02 '19 at 08:13
@eudes I looked at the algorithms but I don't know what can be done about the cubic equation, since I can't say anything about $A$ or $C$ other than what I posted in the question. I think I will resort to numerical approach. — Zaeem Hussain, Mar 07 '19 at 16:09

score 2 · Answer 1 · answered Feb 27 '19 at 19:49

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".

answered Feb 27 '19 at 19:49

Robert Israel

448,999

Dr. Sonnhard Graubner · Accepted Answer · 2019-02-27T19:59:19.210

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.

edited Feb 27 '19 at 19:59

answered Feb 27 '19 at 19:50

Dr. Sonnhard Graubner

95,283

You left out one term. – Robert Israel Feb 27 '19 at 19:51
Yes, thank you, $$A^2$$ was missing – Dr. Sonnhard Graubner Feb 27 '19 at 19:59

score 1 · Answer 3 · answered Feb 27 '19 at 20:40

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.

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

3 Answers3