Whakaoti i te 1+x+x^2=0

$$1 + x + x ^ { 2 } = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{-1+\sqrt{3}i}{2}\approx -0.5+0.866025404i$$

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

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.

$$x^{2}+x+1=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4}}{2}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4}}{2}$$

Tāpiri 1 ki te -4.

$$x=\frac{-1±\sqrt{-3}}{2}$$

Tuhia te pūtakerua o te -3.

$$x=\frac{-1±\sqrt{3}i}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{3}i}{2} ina he tāpiri te ±. Tāpiri -1 ki te i\sqrt{3}.

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

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{3}i}{2} ina he tango te ±. Tango i\sqrt{3} mai i -1.

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

Kua oti te whārite te whakatau.

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

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.

$$x^{2}+x+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x^{2}+x+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$x^{2}+x=-1$$

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.

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

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

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

Tāpiri -1 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{3}{4}$$

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}.

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

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

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

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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