Whakaoti i te 2x^2-2x+5=0

$$2 x ^ { 2 } - 2 x + 5 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{1}{2}+\frac{3}{2}i=0.5+1.5i$$

$$x=\frac{1}{2}-\frac{3}{2}i=0.5-1.5i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$2x^{2}-2x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -2 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\times 5}}{2\times 2}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\times 5}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\times 5}}{2\times 2}$$

Whakareatia -8 ki te 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-40}}{2\times 2}$$

Tāpiri 4 ki te -40.

$$x=\frac{-\left(-2\right)±\sqrt{-36}}{2\times 2}$$

Tuhia te pūtakerua o te -36.

$$x=\frac{-\left(-2\right)±6i}{2\times 2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±6i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{2±6i}{4}$$

Nā, me whakaoti te whārite x=\frac{2±6i}{4} ina he tāpiri te ±. Tāpiri 2 ki te 6i.

$$x=\frac{2+6i}{4}$$

Whakawehe 2+6i ki te 4.

$$x=\frac{1}{2}+\frac{3}{2}i$$

Nā, me whakaoti te whārite x=\frac{2±6i}{4} ina he tango te ±. Tango 6i mai i 2.

$$x=\frac{2-6i}{4}$$

Whakawehe 2-6i ki te 4.

$$x=\frac{1}{2}-\frac{3}{2}i$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$2x^{2}-2x+5=0$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$2x^{2}-2x+5-5=-5$$

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

$$2x^{2}-2x=-5$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-2x}{2}=-\frac{5}{2}$$

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

$$x^{2}+\left(-\frac{2}{2}\right)x=-\frac{5}{2}$$

Whakawehe -2 ki te 2.

$$x^{2}-x=-\frac{5}{2}$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{5}{2}+\left(-\frac{1}{2}\right)^{2}$$

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=-\frac{5}{2}+\frac{1}{4}$$

Tāpiri -\frac{5}{2} ki te \frac{1}{4} 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.

$$x^{2}-x+\frac{1}{4}=-\frac{9}{4}$$

Tauwehea te x^{2}-x+\frac{1}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{3}{2}i x-\frac{1}{2}=-\frac{3}{2}i$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

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