solving $\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$

Question

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?

Answer 1

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 $\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$ $$\to x=1 \space \text{and} \space \sqrt{a}+\sqrt{b}=\sqrt{a+b}$$ $$\to x=1$$

Answer 2

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.