Whakaoti i te x^2=1/2-2sqrt{3}

$${ x }^{ 2 } = \frac{ 1 }{ 2 } -2 \sqrt{ 3 }$$

Whakaoti mō x (complex solution)

$$x=-\frac{i\sqrt{8\sqrt{3}-2}}{2}\approx -0-1.721656648i$$

$$x=\frac{i\sqrt{8\sqrt{3}-2}}{2}\approx 1.721656648i$$

Ngā Upane mā te Kimi Pūtakerua

Kua oti te whārite te whakatau.

$$x=\frac{i\sqrt{8\sqrt{3}-2}}{2} x=-\frac{i\sqrt{8\sqrt{3}-2}}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te \frac{1}{2} mai i ngā taha e rua.

$$x^{2}-\frac{1}{2}=-2\sqrt{3}$$

Me tāpiri te 2\sqrt{3} ki ngā taha e rua.

$$x^{2}-\frac{1}{2}+2\sqrt{3}=0$$

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.

$$x^{2}+2\sqrt{3}-\frac{1}{2}=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(2\sqrt{3}-\frac{1}{2}\right)}}{2}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(2\sqrt{3}-\frac{1}{2}\right)}}{2}$$

Whakareatia -4 ki te -\frac{1}{2}+2\sqrt{3}.

$$x=\frac{0±\sqrt{2-8\sqrt{3}}}{2}$$

Tuhia te pūtakerua o te 2-8\sqrt{3}.

$$x=\frac{0±i\sqrt{8\sqrt{3}-2}}{2}$$

Nā, me whakaoti te whārite x=\frac{0±i\sqrt{8\sqrt{3}-2}}{2} ina he tāpiri te ±.

$$x=\frac{i\sqrt{8\sqrt{3}-2}}{2}$$

Nā, me whakaoti te whārite x=\frac{0±i\sqrt{8\sqrt{3}-2}}{2} ina he tango te ±.

$$x=-\frac{i\sqrt{8\sqrt{3}-2}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{i\sqrt{8\sqrt{3}-2}}{2} x=-\frac{i\sqrt{8\sqrt{3}-2}}{2}$$