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?
Best Answer
This answer proves that $x=1$ is the only real solution.
Proof :
$$(x+1)\sqrt{2x^2 + 2} + \sqrt{6x^2 + 18x +12} = \dfrac{3x^2 + 7x + 10}{2}\tag1$$
First of all, we have to have $$6x^2+18x+12\geqslant 0,$$ i.e. $$x\in(-\infty,-2]\cup [-1,\infty)$$
Case 1 : $x\leqslant -2$
Multiplying LHS of $(1)$ by $\dfrac{\sqrt{6x^2 + 18x +12}-(x+1)\sqrt{2x^2 + 2}}{\sqrt{6x^2 + 18x +12}-(x+1)\sqrt{2x^2 + 2}}\ (=1)$, the equation $(1)$ can be written as $$\frac{6x^2 + 18x +12-(x+1)^2(2x^2 + 2)}{\sqrt{6x^2 + 18x +12}-(x+1)\sqrt{2x^2 + 2}} = \dfrac{3x^2 + 7x + 10}{2}$$ i.e. $$\frac{-2(x + 1)\bigg((x+2)(x-1) x - 5\bigg)}{\sqrt{6x^2 + 18x +12}-(x+1)\sqrt{2x^2 + 2}} = \frac{3}{2}\bigg(x+\frac 76\bigg)^2+\frac{71}{24}$$
The denominator of LHS is positive with $\underbrace{(x+2)(x-1) x}_{\text{non-positive}} - 5\lt 0$, so LHS is negative while RHS is positive, which is a contradiction.
Therefore, our equation $(1)$ does not have any real solution $x$ satisfying $x\leqslant -2$.
Case 2 : $x=\sqrt 3-2$ (if and only if $(x+1)\sqrt{2x^2 + 2} +4x=0$ holds. See $A$ in Case 3.)
This is not a solution.
Case 3 : $x\in [-1,\sqrt 3-2)\cup(\sqrt 3-2,\infty)$
$(1)$ can be written as $$\underbrace{\bigg((x+1)\sqrt{2x^2 + 2} -4x\bigg)}_{A}+ \underbrace{\bigg(\sqrt{6x^2 + 18x +12}-\frac{5x+7}{2}\bigg)}_{B}+\underbrace{\bigg(\frac{13x+7}{2}-\dfrac{3x^2 + 7x + 10}{2}\bigg)}_{C}=0$$
(An explanation : You already know that $x=1$ is a solution. Let $f(x):=(x+1)\sqrt{2x^2 + 2}$. Then, $f(1)=4$. So, $f(x)-4=(x-1)g(x)$ where $g(x)=\dfrac{2(x^3 + 3 x^2 + 5 x + 7)}{(x+1)\sqrt{2x^2 + 2}+4}$. Since $g(1)=4$, one can see that $(x-1)(g(x)-4)=(x-1)g(x)-4(x-1)=f(x)-4-4(x-1)=f(x)-4x$ has a factor $(x-1)^2$. Similarly, one can see that $\sqrt{6x^2 + 18x +12}-\dfrac{5x+7}{2}$ has a factor $(x-1)^2$.)
Now, $A$ can be written as $$\begin{align}A&=\bigg((x+1)\sqrt{2x^2 + 2} -4x\bigg)\cdot\frac{(x+1)\sqrt{2x^2 + 2} +4x}{(x+1)\sqrt{2x^2 + 2} +4x} \\\\&=\frac{(x+1)^2(2x^2+2)-16x^2}{(x+1)\sqrt{2x^2 + 2}+4x} \\\\&=\frac{2 (x^2 + 4 x + 1)(x - 1)^2}{(x+1)\sqrt{2x^2 + 2} +4x}\end{align}$$
$B$ can be written as $$\begin{align}B&=\bigg(\sqrt{6x^2 + 18x +12}-\dfrac{5x+7}{2}\bigg)\cdot\frac{\sqrt{6x^2 + 18x +12}+\dfrac{5x+7}{2}}{\sqrt{6x^2 + 18x +12}+\dfrac{5x+7}{2}} \\\\&=\frac{(6x^2+18x+12)-\bigg(\dfrac{5x+7}{2}\bigg)^2}{\sqrt{6x^2 + 18x +12}+\dfrac{5x+7}{2}} \\\\&=\frac{-\dfrac 14(x - 1)^2}{\sqrt{6x^2 + 18x +12}+\dfrac{5x+7}{2}}\end{align}$$
$C$ can be written as $$C=\frac{13x+7-(3x^2 + 7x + 10)}{2}=\frac{-3(x-1)^2}{2}$$
Therefore, our equation $(1)$ can be written as $$(x-1)^2\underbrace{\bigg(\frac{2(x^2 + 4 x + 1)}{(x+1)\sqrt{2x^2 + 2} +4x}-\frac{1}{4\sqrt{6x^2 + 18x +12}+2(5x+7)}-\frac{3}{2}\bigg)}_{D}=0$$
Finally, let us prove that $D\lt 0$.
$$\begin{align}D&\lt \frac{2(x^2 + 4 x + 1)}{(x+1)\sqrt{2x^2 + 2} +4x}-\frac{3}{2} \\\\&=\frac{4(x^2 + 4 x + 1)-3\bigg((x+1)\sqrt{2x^2 + 2} +4x\bigg)}{2\bigg((x+1)\sqrt{2x^2 + 2} +4x\bigg)} \\\\&=\frac{4(x^2+x+1)-3(x+1)\sqrt{2x^2 + 2}}{2\bigg((x+1)\sqrt{2x^2 + 2} +4x\bigg)} \\\\&=\frac{16(x^2+x+1)^2-9(x+1)^2(2x^2 + 2)}{2\bigg((x+1)\sqrt{2x^2 + 2} +4x\bigg)\bigg(4(x^2+x+1)+3(x+1)\sqrt{2x^2 + 2}\bigg)} \\\\&=\frac{-2 (x - 1)^2 (x^2+ 4x + 1)}{2\bigg((x+1)\sqrt{2x^2 + 2} +4x\bigg)\bigg(4(x^2+x+1)+3(x+1)\sqrt{2x^2 + 2}\bigg)} \\\\&=\underbrace{\frac{-(x-1)^2(x+2+\sqrt 3)}{4(x^2+x+1)+3(x+1)\sqrt{2x^2 + 2}}}_{\text{non-positive}}\cdot\underbrace{\frac{x-(\sqrt 3-2)}{(x+1)\sqrt{2x^2 + 2} +4x}}_{E}\end{align}$$
Here, $E$ is positive since one can prove that $(x+1)\sqrt{2x^2+2}+4x\gt 0$ if and only if $\sqrt 3-2\lt x$ as follows : $$\begin{align}&(x+1)\sqrt{2x^2+2}+4x\gt 0 \\\\&\iff (x+1)\sqrt{2x^2+2}\gt -4x \\\\&\iff x\geqslant 0\ \ \text{or}\ \ \bigg(x\lt 0\ \ \text{and}\ \ (x+1)^2(2x^2+2)\gt (-4x)^2\bigg) \\\\&\iff x\geqslant 0\ \ \text{or}\ \ \bigg(x\lt 0\ \ \text{and}\ \ (x - 1)^2 (x^2 + 4 x + 1)>0\bigg) \\\\&\iff x\geqslant 0\ \ \text{or}\ \ \bigg(x\lt 0\ \ \text{and}\ \ x^2 + 4 x + 1>0\bigg) \\\\&\iff \sqrt 3-2\lt x\end{align}$$
(Since we have already seen in Case 2 that $x=\sqrt 3-2$ if and only if $(x+1)\sqrt{2x^2 + 2} +4x=0$, we can say that $(x+1)\sqrt{2x^2+2}+4x\lt 0$ if and only if $-1\leqslant x\lt\sqrt 3-2$.)
Therefore, we can say that $D$ is negative, which implies that $x=1$ is the only real solution.$\quad\blacksquare$