Whakaoti i te CD=sqrt{left(+8cright)^2-DD^2}

$$CD = \sqrt{ { \left(+8c \right) }^{ 2 } -D { D }^{ 2 } }$$

Whakaoti mō C

$$\left\{\begin{matrix}C=\frac{\sqrt{64c^{2}-D^{3}}}{D}\text{, }&D\neq 0\text{ and }D\leq 4c^{\frac{2}{3}}\\C\in \mathrm{R}\text{, }&D=0\text{ and }c=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 2 kia riro ai te 3.

$$CD=\sqrt{\left(8c\right)^{2}-D^{3}}$$

Whakarohaina te \left(8c\right)^{2}.

$$CD=\sqrt{8^{2}c^{2}-D^{3}}$$

Tātaihia te 8 mā te pū o 2, kia riro ko 64.

$$CD=\sqrt{64c^{2}-D^{3}}$$

He hanga arowhānui tō te whārite.

$$DC=\sqrt{64c^{2}-D^{3}}$$

Whakawehea ngā taha e rua ki te D.

$$\frac{DC}{D}=\frac{\sqrt{64c^{2}-D^{3}}}{D}$$

Mā te whakawehe ki te D ka wetekia te whakareanga ki te D.

$$C=\frac{\sqrt{64c^{2}-D^{3}}}{D}$$