Find the minimum value of $x$ in the given that $\frac{\sqrt{2x^2-1} + \sqrt{x^2-1}}{\sqrt2x^2}=1$

Question

How to simplify the given equation and find the minimum value of $x$ ?
$$\frac{\sqrt{2x^2-1} + \sqrt{x^2-1}}{\sqrt2x^2}=1$$ I do square both sides but I doesn't make any sense

Answer 1

Set $t=x^2$ and clear denominators to get $$\sqrt{2t-1}+\sqrt{t-1}=\sqrt{2}t.$$ Squaring both sides shows that \begin{eqnarray*} 2t^2&=&(2t-1)+2\sqrt{(2t-1)(t-1)}+(t-1)\\ &=&3t-2+2\sqrt{2t^2-3t+1}. \end{eqnarray*} Isolating the square root and squaring again shows that $$(2t^2-3t+2)^2=4(2t^2-3t+1).$$ Rearranging yields the quartic equation $$4t^4-12t^3+9t^2=t^2(2t-3)^2=0.$$ Of course $t=0$ is impossible, so we see that $t=\tfrac32$ and so $x=\sqrt{\tfrac32}$. This shows that the minimum value for $x$ equals $\sqrt{\tfrac32}$.

Answer 2

Put $u = x^2 - 1, v = x^2 \implies \sqrt{u+v} + \sqrt{u} = v\sqrt{2}\implies \sqrt{u+v} = v\sqrt{2} - \sqrt{u}\implies u+v = 2v^2-2\sqrt{2}v\sqrt{u}+u\implies 2v^2 - v - 2\sqrt{2}v\sqrt{u} = 0\implies v(2v-1-2\sqrt{2}\sqrt{u}) = 0\implies v = 0\implies x^2=0 \implies x = 0$ , or $2v= 1+2\sqrt{2}\sqrt{u}\implies 2(u+1) = 1 +2\sqrt{2}\sqrt{u}\implies 2u -2\sqrt{2}\sqrt{u} + 1 = 0\implies (\sqrt{2u} - 1)^2 = 0\implies \sqrt{2u} = 1\implies 2u = 1\implies 2(x^2-1) = 1\implies 2x^2 = 3\implies x = \pm \dfrac{\sqrt{6}}{2}$ . But $x = 0$ is not a solution,so the solutions are : $ x=\pm \dfrac{\sqrt{6}}{2}$.

Answer 3

Let $u=\sqrt{2x^2-1}\ge0,v=\sqrt{x^2-1}\ge0$ $\implies u^2-v^2=x^2$

and $ u+v=\sqrt2(u^2-v^2)$

If $u+v=0,\sqrt{2x^2-1}=\sqrt{x^2-1}=0$ which is untenable

$\implies u-v=\dfrac1{\sqrt2}\implies \sqrt2v=\sqrt2u-1$

and $u^2-2v^2=1\implies u^2-(\sqrt2u-1)^2=1$

$\implies 0=(u-\sqrt2)^2$

$u(\ne0)=?$

$x^2=\dfrac{u^2+1}2=?$