Whakaoti i te b^{2}=3.a^2+c^2+ac=4.b^2+c^2-sqrt{3}ab=7

$$b ^ { 2 } = 3 . a ^ { 2 } + c ^ { 2 } + a c = 4 . b ^ { 2 } + c ^ { 2 } - \sqrt { 3 } a b = 7$$

Whakaoti mō b

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

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

Ngā Upane mā te Kimi Pūtakerua

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

Tangohia te 3 mai i ngā taha e rua.

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

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.

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