Whakaoti i te 2+x^2=5

$$2 + x ^ { 2 } = 5$$

Whakaoti mō x

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

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

Ngā Upane mā te Kimi Pūtakerua

Tangohia te 2 mai i ngā taha e rua.

$$x^{2}=5-2$$

Tangohia te 2 i te 5, ka 3.

$$x^{2}=3$$

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 5 mai i ngā taha e rua.

$$2+x^{2}-5=0$$

Tangohia te 5 i te 2, ka -3.

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

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.

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

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}.

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

Pūrua 0.

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

Whakareatia -4 ki te -3.

$$x=\frac{0±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{0±2\sqrt{3}}{2}$$

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

$$x=\sqrt{3}$$

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

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

Kua oti te whārite te whakatau.

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