<s>[INST] 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?
 [/INST] Let $x+1=y, \;y\geqslant 1$, then you have:
$$
\begin{align}\sqrt y+\sqrt {y-1}+\sqrt {y+1}=\sqrt {y+6}\end{align}
$$
Dividing both side of the equation by $\sqrt y$ and using the substitution $\dfrac 1y=u,\;u>0$, then you obtain:
$$
\begin{align}&\sqrt {1-u}+\sqrt {1+u}=\sqrt {1+6u}-1\\
\implies &\sqrt {1-u^2}=3u-\sqrt {1+6u}\\
\implies &10u^2+6u=6u\sqrt {1+6u}\\
\implies &3\sqrt {6u+1}=5u+3\\
\implies &9(6u+1)=(5u+3)^2\\
\implies &u=\frac {24}{25}\\
\implies &x=\frac {1}{24}.\end{align}
$$
</s>