<s>[INST] 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?
 [/INST] 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.
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