Whakaoti i te 4+sqrt{2}x^2=2

$$4+ \sqrt{ 2 } { x }^{ 2 } =2$$

Whakaoti mō x (complex solution)

$$x=-\sqrt[4]{2}i\approx -0-1.189207115i$$

$$x=\sqrt[4]{2}i\approx 1.189207115i$$

Ngā Upane mā te Kimi Pūtakerua

Tangohia te 4 mai i ngā taha e rua.

$$\sqrt{2}x^{2}=2-4$$

Tangohia te 4 i te 2, ka -2.

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

Mā te whakawehe ki te \sqrt{2} ka wetekia te whakareanga ki te \sqrt{2}.

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

Whakawehe -2 ki te \sqrt{2}.

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

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

$$x=\sqrt[4]{2}i x=-\sqrt[4]{2}i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 2 mai i ngā taha e rua.

$$4+\sqrt{2}x^{2}-2=0$$

Tangohia te 2 i te 4, ka 2.

$$2+\sqrt{2}x^{2}=0$$

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.

$$\sqrt{2}x^{2}+2=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\sqrt{2}\times 2}}{2\sqrt{2}}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\sqrt{2}\times 2}}{2\sqrt{2}}$$

Whakareatia -4 ki te \sqrt{2}.

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

Whakareatia -4\sqrt{2} ki te 2.

$$x=\frac{0±\sqrt{-8\sqrt{2}}}{2\sqrt{2}}$$

Tuhia te pūtakerua o te -8\sqrt{2}.

$$x=\frac{0±2\times 2^{\frac{3}{4}}i}{2\sqrt{2}}$$

Nā, me whakaoti te whārite x=\frac{0±2\times 2^{\frac{3}{4}}i}{2\sqrt{2}} ina he tāpiri te ±.

$$x=\frac{2i}{2^{\frac{3}{4}}}$$

Nā, me whakaoti te whārite x=\frac{0±2\times 2^{\frac{3}{4}}i}{2\sqrt{2}} ina he tango te ±.

$$x=-\sqrt[4]{2}i$$

Kua oti te whārite te whakatau.

$$x=\frac{2i}{2^{\frac{3}{4}}} x=-\sqrt[4]{2}i$$