solve an implicit function

Question

I have seen that an example of implicit function that can be solved only numerically is (solving for x knowing y): $$ \sin(x) = y\cdot x $$ I was wondering if the following function can also be solved only numerically: $$ x=\sqrt{x+1} + y $$ For me would be helpful to find a closed form solution, maybe using Mathematica, WxMaxima or any other software.

Thank you.

Just to add the maxima code: maxima workspace

While this is the equation I am trying to solve for $\beta$ (for which I thought that the problem was the root): $$ -((2*\beta^2+2)*d_x*m_x*\sqrt{(1-\theta^2)}+((\beta^2+1)*m_x^2+(-\beta^2-1)*m_x^2)*\theta^2+\sqrt{(\beta^2+1)}*((2*m_x*rho_1-2*m_x*p_z+2*\beta*d_y*m_x)*\theta-2*\beta*m_y*rho_1+2*\beta*m_y*p_z+2*d_y*m_y)+ (-\beta^2-1)*rho_1^2+(2*\beta^2+2)*p_z*rho_1+(-\beta^2-1)*p_z^2+(-\beta^2-1)*m_x^2+(-\beta^2-1)*m_y^2+(\beta^2+1)*L_r^2+(-\beta^2-1)*d_y^2+(-\beta^2-1)*d_x^2)/(\beta^2+1)=0 $$

Answer 1

To get rid of the square root, let $x+1=t^2$, with $t\ge0$. The equation now reads

$$t^2-t-y-1=0.$$

It has the solutions

$$t=\frac{1\pm\sqrt{1+4(y+1)}}2$$ which are defined for $4y+5\ge0$.

We have $t\ge0$ with the plus sign, or with the minus sign and $y\le-1$.

For these values,

$$x=t^2-1=\left(\frac{1\pm\sqrt{1+4(y+1)}}2\right)^2-1.$$

Answer 2

Doing what the comments advice:

$$x^2-2xy+y^2=x+1\implies x^2-(2y+1)x+y^2-1=0$$

and this is a quadratic in $\;x\;$ :

$$\Delta=(2y+1)^2-4(y^2-1)=4y+5\implies x_{1,2}=\frac{(2y+1)\pm\sqrt\Delta}{2}$$

After that, substitute in the original equation and check what fits in.