Whakaoti i te (pip/T_{b})^2=8

$$( \frac { \pi p } { T _ { b } } ) ^ { 2 } = 8$$

Whakaoti mō T_b (complex solution)

$$T_{b}=\frac{\pi \sqrt{2}p}{4}$$

$$T_{b}=-\frac{\pi \sqrt{2}p}{4}\text{, }p\neq 0$$

Whakaoti mō p (complex solution)

$$p=\frac{2\sqrt{2}T_{b}}{\pi }$$

$$p=-\frac{2\sqrt{2}T_{b}}{\pi }\text{, }T_{b}\neq 0$$

Whakaoti mō T_b

$$T_{b}=-\frac{\pi \sqrt{2}p}{4}$$

$$T_{b}=\frac{\pi \sqrt{2}p}{4}\text{, }p\neq 0$$

Whakaoti mō p

$$p=-\frac{2\sqrt{2}T_{b}}{\pi }$$

$$p=\frac{2\sqrt{2}T_{b}}{\pi }\text{, }T_{b}\neq 0$$