Whakaoti i te 2x^2-4x+12=0

$$2 x ^ { 2 } - 4 x + 12 = 0$$

Whakaoti mō x (complex solution)

$$x=1+\sqrt{5}i\approx 1+2.236067977i$$

$$x=-\sqrt{5}i+1\approx 1-2.236067977i$$

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.

$$2x^{2}-4x+12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -4 mō b, me 12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 12}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 12}}{2\times 2}$$

Whakareatia -8 ki te 12.

$$x=\frac{-\left(-4\right)±\sqrt{16-96}}{2\times 2}$$

Tāpiri 16 ki te -96.

$$x=\frac{-\left(-4\right)±\sqrt{-80}}{2\times 2}$$

Tuhia te pūtakerua o te -80.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±4\sqrt{5}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{4±4\sqrt{5}i}{4}$$

Nā, me whakaoti te whārite x=\frac{4±4\sqrt{5}i}{4} ina he tāpiri te ±. Tāpiri 4 ki te 4i\sqrt{5}.

$$x=\frac{4+4\sqrt{5}i}{4}$$

Whakawehe 4+4i\sqrt{5} ki te 4.

$$x=1+\sqrt{5}i$$

Nā, me whakaoti te whārite x=\frac{4±4\sqrt{5}i}{4} ina he tango te ±. Tango 4i\sqrt{5} mai i 4.

$$x=\frac{-4\sqrt{5}i+4}{4}$$

Whakawehe 4-4i\sqrt{5} ki te 4.

$$x=-\sqrt{5}i+1$$

Kua oti te whārite te whakatau.

$$x=1+\sqrt{5}i x=-\sqrt{5}i+1$$

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.

$$2x^{2}-4x+12=0$$

Me tango 12 mai i ngā taha e rua o te whārite.

$$2x^{2}-4x+12-12=-12$$

Mā te tango i te 12 i a ia ake anō ka toe ko te 0.

$$2x^{2}-4x=-12$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-4x}{2}=-\frac{12}{2}$$

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

$$x^{2}+\left(-\frac{4}{2}\right)x=-\frac{12}{2}$$

Whakawehe -4 ki te 2.

$$x^{2}-2x=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$x^{2}-2x=-6$$

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.

$$x^{2}-2x+1=-6+1$$

Tāpiri -6 ki te 1.

$$x^{2}-2x+1=-5$$

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

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

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

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

Whakarūnātia.

$$x-1=\sqrt{5}i x-1=-\sqrt{5}i$$

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

$$x=1+\sqrt{5}i x=-\sqrt{5}i+1$$