समाधान sqrt[4x]{y^2+3}=x^x

$$\sqrt[ 4x ]{ { y }^{ 2 } +3 } = { x }^{ x }$$

y को लागि हल गर्नुहोस् (complex solution)

$$y=e^{4\left(Im(\frac{1}{x})+iRe(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)arg(x^{x})-8\pi Re(\frac{1}{x})n_{1}i\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)-8\pi Im(\frac{1}{x})n_{1}\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}\left(|x^{x}|\right)^{4\left(Re(\frac{1}{x})-iIm(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}$$

$$n_{1}\in \mathrm{Z}$$

$$x\neq 0$$

y को लागि हल गर्नुहोस्

$$y=\sqrt{\left(x^{x}\right)^{4x}-3}$$

$$y=-\sqrt{\left(x^{x}\right)^{4x}-3}\text{, }\left(x<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-x\right)^{x^{2}}\geq \sqrt[4]{3}\text{ and }x^{x}\geq 0\right)\text{ or }\left(x<0\text{ and }Denominator(x)\text{bmod}2=1\text{ and }\left(-x\right)^{x^{2}}\geq \sqrt[4]{3}\text{ and }Numerator(\frac{1}{4x})\text{bmod}2=1\text{ and }Denominator(\frac{1}{4x})\text{bmod}2=1\right)\text{ or }\left(x>0\text{ and }x^{x^{2}}\geq \sqrt[4]{3}\text{ and }Numerator(\frac{1}{4x})\text{bmod}2=1\text{ and }Denominator(\frac{1}{4x})\text{bmod}2=1\right)\text{ or }\left(x>0\text{ and }x^{x^{2}}\geq \sqrt[4]{3}\text{ and }x^{x}\geq 0\right)$$