Whakaoti i te 2x^2+1/2=x

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

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{3}i}{4}\approx 0.25+0.433012702i$$

$$x=\frac{-\sqrt{3}i+1}{4}\approx 0.25-0.433012702i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x mai i ngā taha e rua.

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

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

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

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

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

Whakareatia -4 ki te 2.

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

Whakareatia -8 ki te \frac{1}{2}.

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

Tāpiri 1 ki te -4.

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

Tuhia te pūtakerua o te -3.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{3}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{1±\sqrt{3}i}{4}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{3}i}{4} ina he tāpiri te ±. Tāpiri 1 ki te i\sqrt{3}.

$$x=\frac{1+\sqrt{3}i}{4}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{3}i}{4} ina he tango te ±. Tango i\sqrt{3} mai i 1.

$$x=\frac{-\sqrt{3}i+1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{1+\sqrt{3}i}{4} x=\frac{-\sqrt{3}i+1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x mai i ngā taha e rua.

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

Tangohia te \frac{1}{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe -\frac{1}{2} ki te 2.

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

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.

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

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{4}+\frac{1}{16}$$

Tāpiri -\frac{1}{4} ki te \frac{1}{16} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{3}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{3}{16}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{3}{16}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{\sqrt{3}i}{4} x-\frac{1}{4}=-\frac{\sqrt{3}i}{4}$$

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

$$x=\frac{1+\sqrt{3}i}{4} x=\frac{-\sqrt{3}i+1}{4}$$