Whakaoti i te x^2=pi

$${ x }^{ 2 } = \pi$$

Whakaoti mō x

$$x=\sqrt{\pi }\approx 1.772453851$$

$$x=-\sqrt{\pi }\approx -1.772453851$$

Ngā Upane mā te Kimi Pūtakerua

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$x^{2}=\pi$$

Me tango \pi mai i ngā taha e rua o te whārite.

$$x^{2}-\pi =\pi -\pi$$

Mā te tango i te \pi i a ia ake anō ka toe ko te 0.

$$x^{2}-\pi =0$$

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

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

Pūrua 0.

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

Whakareatia -4 ki te -\pi .

$$x=\frac{0±\sqrt{4\pi }}{2}$$

Tuhia te pūtakerua o te 4\pi .

$$x=\frac{0±2\sqrt{\pi }}{2}$$

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

$$x=\sqrt{\pi }$$

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

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

Kua oti te whārite te whakatau.

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