Whakaoti i te 1^2+b^2=2^2

$$1 ^ { 2 } + b ^ { 2 } = 2 ^ { 2 }$$

Whakaoti mō b

$$b=\sqrt{3}\approx 1.732050808$$

$$b=-\sqrt{3}\approx -1.732050808$$

Ngā Upane mā te Kimi Pūtakerua

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$1+b^{2}=2^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$1+b^{2}=4$$

Tangohia te 1 mai i ngā taha e rua.

$$b^{2}=4-1$$

Tangohia te 1 i te 4, ka 3.

$$b^{2}=3$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$b=\sqrt{3} b=-\sqrt{3}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$1+b^{2}=2^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$1+b^{2}=4$$

Tangohia te 4 mai i ngā taha e rua.

$$1+b^{2}-4=0$$

Tangohia te 4 i te 1, ka -3.

$$-3+b^{2}=0$$

Ko ngā tikanga tātai pūrua pēnei i tēnei nā, me te kīanga tau x^{2} engari kāore he kīanga tau x, ka taea tonu te whakaoti mā te whakamahi i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, ina tuhia ki te tānga ngahuru: ax^{2}+bx+c=0.

$$b^{2}-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Pūrua 0.

$$b=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Whakareatia -4 ki te -3.

$$b=\frac{0±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$b=\frac{0±2\sqrt{3}}{2}$$

Nā, me whakaoti te whārite b=\frac{0±2\sqrt{3}}{2} ina he tāpiri te ±.

$$b=\sqrt{3}$$

Nā, me whakaoti te whārite b=\frac{0±2\sqrt{3}}{2} ina he tango te ±.

$$b=-\sqrt{3}$$

Kua oti te whārite te whakatau.

$$b=\sqrt{3} b=-\sqrt{3}$$