Using "symmetry" to deduce about two colliding particles

[brotherbobby]

TL;DR

A textbook says this : "... if a particle of mass travels at a velocity and undergoes an elastic collision with another initially stationary particle of mass , the expression for the final velocities of the particles should not be symmetric."

I am struggling to understand this. Additionally, the statement above makes just as little sense when I carry out the mathematics. I write below the details, realising that space is brief here.

Text excerpt : For context, I copy and paste the paragraph on "symmetry" from the text, underlining the relevant lines in red. I hope it's readable.

Problem situation : This concerns the last three lines from the text excerpt. They say that :

"... if a particle of mass $m_1$ travels at a velocity $u$ and undergoes an elastic collision with another initially stationary particle of mass $m_2$, the expression for the final velocities of the particles should not be symmetric."

Doubt(s) : I have several.

1. What does it even mean for the final velocities to be symmetric ? Does it mean that if we replace $m_1\rightarrow m_2$ in the expression for the velocity for $m_1$, we will obtain the expression for the velocity for $m_2$?
2. Why should the expression(s) for the final velocities be not symmetric?

Attempt : I attempt the problem in two parts, tackling the symmetric case first.

(The symmetry argument) : The particle of mass $m_1$ moving with velocity $u$ to the right relative to particle of mass $m_2$ is the same as the latter moving with velocity $-u$ to the left relative to the first. But the second particle has a different mass $m_2$. The two situations are not symmetric before collision, hence the final velocities will also not be following collision.


(The mathematics) : From the picture drawn for the problem, momentum conservation yields $\small{m_1 u=m_1v_1+m_2v_2\Rightarrow v_2=\dfrac{m_1}{m_2}(u-v_1)\quad\color{blue}{\dots 1}}$. From energy conservation : $\small{m_1 u^2=m_1v_1^2+m_2v_2^2=m_1v_1^2+\dfrac{m_1^2}{m_2^2}(u-v_1)^2\Rightarrow u^2=v_1^2+\dfrac{m_1}{m_2}(u-v_1)^2\Rightarrow \left(1+\dfrac{m_1}{m_2}\right)v_1^2-\dfrac{2m_1u}{m_2}v_1-u^2\left(1-\dfrac{m_1}{m_2}\right)=0}$, which can be solved to give :
$\small{v_1=\dfrac{\dfrac{2m_1u}{m_2}\pm\sqrt{\dfrac{4m_1^2u^2}{m_2^2}+4u^2\left( 1- \dfrac{m_1^2}{m_2^2}u^2\right)}}{2\left(1+\dfrac{m_1}{m_2}\right)}=\dfrac{\dfrac{m_1}{m_2}u\pm u}{1+\dfrac{m_1}{m_2}}=u\;,\; \dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}u}$.

So the final velocity of the first particle can be $\boxed{(v_1=)\,u}$. Using $\small{\color{blue}{1}, \boxed{v_2=0}}$, the final velocity of the second particle. If I am correct, then indeed no symmetry exists in the final velocities.

Or if the final velocity of the first particle $\small{\boxed{(v_1=)\,\dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}u}}\Rightarrow \boxed{v_2}=\dfrac{m_1}{m_2}u\left(1-\dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}\right)=\dfrac{m_1}{m_2}u\dfrac{2}{1+\dfrac{m_1}{m_2}}=\boxed{\dfrac{m_1}{m_2}\dfrac{2u}{1+\dfrac{m_1}{m_2}}=v_2}$, which is the final velocity of the second particle, again showing no symmetry exists.

Request : Am I correct in my arguments above?