Whakaoti i te 2x^2=x-4

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

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{31}i}{4}\approx 0.25+1.391941091i$$

$$x=\frac{-\sqrt{31}i+1}{4}\approx 0.25-1.391941091i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x mai i ngā taha e rua.

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

Me tāpiri te 4 ki ngā taha e rua.

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

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

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 4}}{2\times 2}$$

Whakareatia -8 ki te 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-32}}{2\times 2}$$

Tāpiri 1 ki te -32.

$$x=\frac{-\left(-1\right)±\sqrt{-31}}{2\times 2}$$

Tuhia te pūtakerua o te -31.

$$x=\frac{-\left(-1\right)±\sqrt{31}i}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{31}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{1±\sqrt{31}i}{4}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{31}i}{4} ina he tāpiri te ±. Tāpiri 1 ki te i\sqrt{31}.

$$x=\frac{1+\sqrt{31}i}{4}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{31}i}{4} ina he tango te ±. Tango i\sqrt{31} mai i 1.

$$x=\frac{-\sqrt{31}i+1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{1+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i+1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x mai i ngā taha e rua.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}-\frac{1}{2}x=-\frac{4}{2}$$

Whakawehe -4 ki te 2.

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

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-2+\left(-\frac{1}{4}\right)^{2}$$

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-2+\frac{1}{16}$$

Tāpiri -2 ki te \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{31}{16}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{31}{16}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{\sqrt{31}i}{4} x-\frac{1}{4}=-\frac{\sqrt{31}i}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{1+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i+1}{4}$$