<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] There is a fairly simple approach to this:
$\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+7}$
$\sqrt{x}+\sqrt{x+1}=\sqrt{x+7}-\sqrt{x+2}$
Now, squaring both sides, we get:
$2x+1+2\sqrt{x(x+1)}=2x+9-2\sqrt{(x+2)(x+7)}$
$\sqrt{x(x+1)}=4-\sqrt{(x+2)(x+7)}$
Squaring once again:
$x(x+1)=16-8\sqrt{(x+2)(x+7)}+(x+2)(x+7)$
$x^2+x=16-8\sqrt{(x+2)(x+7)}+x^2+9x+14$
$8\sqrt{(x+2)(x+7)}=30+8x$
$4\sqrt{(x+2)(x+7)}=15+4x$
Squaring once again:
$16(x+2)(x+7)=225+120x+16x^2$
$16(x^2+9x+14)=225+120x+16x^2$
From here, the solution is very straight forward with a simple quadratic equation that you can solve easily. Make sure to verify the roots you get by plugging them in and checking if they satisfy the original equation. Some of them could be extraneous due to squaring of both sides.
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