Solving $(x+1)\sqrt{2(x^2 + 1)} + \sqrt{6x^2 + 18x +12} = \frac{3x^2 + 7x + 10}{2}$

Question

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?

Answer 1

Part 1: real solution

From $6x^2 + 18x + 12 = 6(x + 1)(x + 2) \ge 0$, we have $x \le -2$ or $x \ge -1$.

1. $x \le -2$:

We have $$6x^2 + 18x + 12 = 6(x + 1)(x + 2) \le 6(x + 1)^2 < 9(x + 1)^2.$$ Thus, we have $$\sqrt{6x^2 + 18x + 12} < \sqrt{9(x + 1)^2} = -3(x + 1).$$ Using $(x + 1)\sqrt{2(x^2 + 1)} < 0$, we have $$\mathrm{RHS} - \mathrm{LHS} > \frac{3x^2 + 7x + 10}{2} + 3(x + 1) = \frac{3x^2 + 13x + 16}{2} > 0.$$ Thus, no real solution.

2. $x \ge -1$:

We have $$\left(\frac{7 + 5x}{2}\right)^2 - (6x^2 + 18x + 12) = \frac14(x - 1)^2 \ge 0.$$ Remark: $\frac{7 + 5x}{2}$ is the first order Taylor approximation of $\sqrt{6x^2 + 18x + 12}$ around $x = 1$.

Thus, we have $$\frac{7 + 5x}{2} \ge\sqrt{6x^2 + 18x + 12}. $$ We have \begin{align*} \mathrm{RHS} - \mathrm{LHS} &\ge \frac{3x^2 + 7x + 10}{2} - \frac{7 + 5x}{2} - (x + 1)\sqrt{2(1 + x^2)}\\ &= \frac{3x^2 + 2x + 3}{2} - (x + 1)\sqrt{2(1 + x^2)}\\ &= \frac{(x - 1)^4/4}{(3x^2 + 2x + 3)/2 + (x + 1)\sqrt{2(1 + x^2)}}\\ &\ge 0 \end{align*} with equality if and only if $x = 1$.

Thus, $x = 1$ is the unique real solution.

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$\phantom{2}$

Part 2: non-real complex solution

Perhaps we have to do something similar to that Donald Splutterwit has done. With the help of CAS (Computer Algebra System), we have $$(x^6 + 54x^5 + 153x^4 + 104x^3 - 72x^2 + 400)(x - 1)^2 = 0.$$ Consider the equation $$x^6 + 54x^5 + 153x^4 + 104x^3 - 72x^2 + 400 = 0. \tag{1}$$ Using SageMath, the roots of (1) cannot be expressed using radicals.

(1) has exactly two real roots $x_1 \approx -51.04298389, x_2 \approx -2.074797947$ which are discarded directly. We need to check each non-real complex root of (1).

Answer 2

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.

Answer 3

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.$\quad\blacksquare$

Answer 4

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

Answer 5

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.