ແກ້ mufg=b^2

$$\mu f g = b ^ { 2 }$$

ແກ້ສຳລັບ f (complex solution)

$$\left\{\begin{matrix}f=\frac{b^{2}}{g\mu }\text{, }&g\neq 0\text{ and }\mu \neq 0\\f\in \mathrm{C}\text{, }&\left(g=0\text{ or }\mu =0\right)\text{ and }b=0\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$g\mu f=b^{2}$$

ຫານທັງສອງຂ້າງດ້ວຍ \mu g.

$$\frac{g\mu f}{g\mu }=\frac{b^{2}}{g\mu }$$

ການຫານດ້ວຍ \mu g ຈະຍົກເລີກການຄູນດ້ວຍ \mu g.

$$f=\frac{b^{2}}{g\mu }$$

ແກ້ສຳລັບ f

$$\left\{\begin{matrix}f=\frac{b^{2}}{g\mu }\text{, }&g\neq 0\text{ and }\mu \neq 0\\f\in \mathrm{R}\text{, }&\left(g=0\text{ or }\mu =0\right)\text{ and }b=0\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$g\mu f=b^{2}$$

ຫານທັງສອງຂ້າງດ້ວຍ \mu g.

$$\frac{g\mu f}{g\mu }=\frac{b^{2}}{g\mu }$$

ການຫານດ້ວຍ \mu g ຈະຍົກເລີກການຄູນດ້ວຍ \mu g.

$$f=\frac{b^{2}}{g\mu }$$

ແກ້ສຳລັບ b (complex solution)

$$b=-\sqrt{f}\sqrt{g}\sqrt{\mu }$$

$$b=\sqrt{f}\sqrt{g}\sqrt{\mu }$$

ແກ້ສຳລັບ b

$$b=\sqrt{fg\mu }$$

$$b=-\sqrt{fg\mu }\text{, }\left(g\leq 0\text{ and }f\leq 0\text{ and }\mu \geq 0\right)\text{ or }\left(f\geq 0\text{ and }g\geq 0\text{ and }\mu \geq 0\right)\text{ or }\left(\mu \leq 0\text{ and }f\leq 0\text{ and }g\geq 0\right)\text{ or }\left(\mu \leq 0\text{ and }g\leq 0\text{ and }f\geq 0\right)$$