Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$ Today, I came across this problem.
$$(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x  +12} = \dfrac{3x^2 + 7x + 10}{2}$$
We are asked to find the possible values of $x$ satisfying this equation.
The first thought which came to my mind is to use some kind of factorisation. I tried for like an hour but all in vain.
Then, I tried to solve the problem by multiplying both sides by 2 and then squaring both sides. The equation became too complicated.
By using some hit and trial, I get to know that $x = 1$ satisfies the equation. But what about complex solutions. So this method is also of no use.
I am sure this question has to be solved using some special equality which I'm unaware of. I want a method so that, I could obtain all the possible values. Can anyone help me or just give some hints?
 A: Square the equation rearrange & square again. This will allow you to remove the radicals & leave you with a polynomial.
\begin{eqnarray*}
(3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12)=8(x+1)\sqrt{2(x^2+1)(6x^2+18x+12)} \\
\end{eqnarray*}
\begin{eqnarray*}
 ((3x^2+7x+10)^2-8(x+1)^2(x^2+1)-4(6x^2+18x+12))^2-128(x+1)^2(x^2+1)(6x^2+18x+12)=0.
\end{eqnarray*}
Now I would resort to CAS ...

So $x=1$ is the only real solution as $x^6+54x^5+153x^4+104x^3 -72x^2+400$ is always positive. The complex roots will be jolly unpleasant & unilluming.
A: Let $x> -1$ and denotes by :
$$f\left(x\right)=(x+1)\sqrt{2(x^{2}+1)}+\sqrt{6x^{2}+18x+12}-\frac{3x^{2}+7x+10}{2}$$
We have :
$$f''(x)=-\frac{\sqrt{\frac{3}{2}}}{2\left(x^{2}+3x+2\right)^{\frac{3}{2}}}+\frac{\sqrt{2}\left(2x^{3}+3x+1\right)}{\left(x^{2}+1\right)^{\frac{3}{2}}}-3$$
It's not hard to prove that :
$$-\frac{\sqrt{\frac{3}{2}}}{2\left(x^{2}+3x+2\right)^{\frac{3}{2}}}\leq 0$$
And :
$$\frac{\sqrt{2}\left(2x^{3}+3x+1\right)}{\left(x^{2}+1\right)^{\frac{3}{2}}}-3\leq 0$$
Therefore on his domain the first derivative is decreasing where $f'(1)=0$. But $f(1)=0$
It shows that  the unique solution is $x=1$
Edit :
With the Bagis's works and  Donald Splutterwit's answer we can find the other roots in term of nested radicals see https://www.researchgate.net/publication/262973394_Solution_of_Polynomial_Equations_with_Nested_Radicals
A: There is only a single real solution:
For $x\geq -1$ :
$$\left[\sqrt{3} \sqrt{1+x} - \sqrt{2} \sqrt{2+x}\right]^2 + \left[(1+x) - \sqrt{2} \sqrt{1+x^2}\right]^2 \geq 0$$
$$\left[7 + 5 x - 2 \sqrt{6(1+x)(2+x)} \right] + \left[ 3 + 2 x + 3 x^2 - 2 (1+x) \sqrt{2(1+x^2)} \right] \geq 0$$
$$ 10 + 7x + 3 x^2 \geq 2 \sqrt{6(1+x)(2+x)} + 2 (1+x) \sqrt{2(1+x^2)}$$
Both terms in the first line are zero if and only if $x=1$.
For $x\leq -2$ additional minus signs are required within both square roots of the first term, but no additional solution exists.
For $-2<x<-1$ the term $\sqrt{6(1+x)(2+x)}$ results in complex numbers.
