solving $\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$ I am trying to solve this equation
$$\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$$
By using Mathematica, I know that, the equation has unique solution  $x=1$.
I tried to write the equation in the form
$$\sqrt{(x-1) (x+1) \left(2x^2-1\right)} + \sqrt{x^2\left(2x^2-1\right)} =4 x-3.$$
From here, I can not solve it. How can I solve it?

Note that the domain of the equation is $x\ge1$ and
$$\sqrt{2x^4-x^2} - (2x^2-1)= \sqrt{2x^2-1} \left( x-\sqrt{2x^2-1}\right)
= -\frac{\sqrt{2x^2-1} (x^2-1)}{x+\sqrt{2x^2-1} }
$$
Then, factorize the equation as follows
\begin{align}
&\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}-(4 x-3)\\
=& \sqrt{(x^2-1) (2x^2-1)} +\left( \sqrt{x^2(2x^2-1)} - (2x^2-1)\right)+2(x^2-1)^2\\
 =& \sqrt{(x^2-1) (2x^2-1)} -\frac{\sqrt{2x^2-1} (x^2-1)}{x+\sqrt{2x^2-1} } +2(x^2-1)^2 \\
 =& \sqrt{(x^2-1) (2x^2-1)} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2(x^2-1)^2 \\
 =& \sqrt{(x^2-1) (2x^2-1)} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2(x^2-1)^2\\
=& \sqrt{x^2-1}\cdot f(x)
\end{align}
where
$$f(x) = \sqrt{2x^2-1} \> \frac{x+\sqrt{2x^2-1}-\sqrt{x^2-1}}{x+\sqrt{2x^2-1} } +2\sqrt{x^2-1}>0
$$
Thus, $x=1$ is the sole solution.
