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?

let $a=2x^4-3x^2+1,b=2x^4-x^2$ where $ a,b\ge 0$ now $\sqrt{a+b}=2x^2-1$ so $$\sqrt{a}+\sqrt{b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+\sqrt{a+b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+2{(x-1)}^2=0$$ because $\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$ $$\to x=1 \space  \text{and} \space \sqrt{a}+\sqrt{b}=\sqrt{a+b}$$ $$\to x=1$$
