Whakaoti i te x^2-2x+3=0

$$x ^ { 2 } - 2 x + 3 = 0$$

Whakaoti mō x (complex solution)

$$x=1+\sqrt{2}i\approx 1+1.414213562i$$

$$x=-\sqrt{2}i+1\approx 1-1.414213562i$$

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.

$$x^{2}-2x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -2 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12}}{2}$$

Tāpiri 4 ki te -12.

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

Tuhia te pūtakerua o te -8.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{2}i}{2}$$

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

$$x=\frac{2+2\sqrt{2}i}{2}$$

Whakawehe 2+2i\sqrt{2} ki te 2.

$$x=1+\sqrt{2}i$$

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

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

Whakawehe 2-2i\sqrt{2} ki te 2.

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

Kua oti te whārite te whakatau.

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

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.

$$x^{2}-2x+3=0$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$x^{2}-2x+3-3=-3$$

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

$$x^{2}-2x=-3$$

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

$$x^{2}-2x+1=-3+1$$

Tāpiri -3 ki te 1.

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

Tauwehea x^{2}-2x+1. 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}.

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

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

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

Whakarūnātia.

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

Me tāpiri 1 ki ngā taha e rua o te whārite.

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