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

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

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.

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

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

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\times 2}}{2\times 2}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\times 2\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-2±\sqrt{4-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

$$x=\frac{-2±\sqrt{4-16}}{2\times 2}$$

Tāpiri 4 ki te -16.

$$x=\frac{-2±\sqrt{-12}}{2\times 2}$$

Tuhia te pūtakerua o te -12.

$$x=\frac{-2±2\sqrt{3}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-2±2\sqrt{3}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{3}i}{4} ina he tāpiri te ±. Tāpiri -2 ki te 2i\sqrt{3}.

$$x=\frac{-2+2\sqrt{3}i}{4}$$

Whakawehe -2+2i\sqrt{3} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{3}i}{4} ina he tango te ±. Tango 2i\sqrt{3} mai i -2.

$$x=\frac{-2\sqrt{3}i-2}{4}$$

Whakawehe -2-2i\sqrt{3} ki te 4.

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

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.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 2 ki te 2.

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

Whakawehe -2 ki te 2.

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

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 x^{2}+x+\frac{1}{4}. 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}.

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

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.

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