Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$? I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2}  = \sqrt{x+7}.$
Answer is $\frac1{24}$.
The first equation has two solutions however the latter one has only one solution. I lossed up one solution but still, I'm interested in solving it.
I wish if someone could help me in solving any of the above equations.
Here's my work:
$$\begin{align}x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} &= 2\\x + \frac{1}{2}\left(2 \sqrt{x}\sqrt{x+1} + 2\sqrt{x+1}\sqrt{x+2} + 2\sqrt{x}\sqrt{x+2}\right) &=2\tag{1}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (x + x + 1 + x + 2)\Big]& = 2\tag{2}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 2\\2x + \Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 4\\(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 & = x + 7\\\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} & = \sqrt{x+7}\end{align}$$
Moving from $(1)$ to $(2)$, I used $2(ab + bc + ca) = (a+b+c)^2 - (a^2+b^2+c^2)$.

In general, is it possible to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}  = \sqrt{x+k}$ by hand?

Given the equation
$$ \sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7} $$
consider it in the form
$$ \sqrt{x} + \sqrt{x+1} = \sqrt{x+7} - \sqrt{x+2} $$
then by squaring both sides the result is
$$ \sqrt{x (x+1)} = 4 - \sqrt{(x+2)(x+7)}. $$
Now, with this equation written as $4 = \sqrt{x (x+1)} + \sqrt{(x+2)(x+7)}$ then, by squaring both sides,
$$ \sqrt{x (x+1) (x+2) (x+7)} = x^2 + 5 x -1. $$
Again squaring both sides leads to
$$ x (x+1) (x+2)(x+7) = x^4 + 10 x^3 + 23 x^2 - 10 x + 1 = x(x+1)(x+2)(x+7) - 24 x + 1$$
which gives the equation $$24 x = 1 \hspace{5mm} \text{and the result} \, x = \frac{1}{24}. $$
