Finding real solutions to $\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2$

Question

I found that $x=1$ is a solution of $$\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}=x^2+2.$$

How would one find other real solutions?

Answer 1

As Dr. Sonnhard Graubner commented, the problem is to find the second $x$ intercept of function $$f(x)=\sqrt{x}+\sqrt[3]{x^2-1}+\sqrt[4]{x^3+15}-x^2-2$$ Such a monster will not show any analytical solution and only numerical methods (such as Newton) will do the job.

I suppose that, from a plot of the function, you noticed that the root is close to $x_0=1.5$. So, using this value as a starting point and applying Newton method, the iterates would be $$x_1=1.579472329$$ $$x_2=1.574652774$$ $$x_3=1.574635699$$ which is the solution for ten significant figures.

Answer 2

To find a polynomial with that root:
You have $A+B+C=x^2+2$. Consider $(x^2+2)^2=(A+B+C)^2= x+B^2+C^2+2AB+2AC+2BC$.
Take the first 25 powers of $x^2+2$, from $1=(x^2+2)^0$ to $(x^2+2)^{24}$. They will all be linear combinations of $A,B,C,B^2,C^2,C^3,AB,AB^2,AC,ABC,AB^2C,...$ with coefficients that are polynomials in $x$. But that is a 24-dimensional space, so there is a linear relation between the 25 formulas. This linear relation is a polynomial in $x$.