Whakaoti i te 3x^2+2x+1=0

$$3 x ^ { 2 } + 2 x + 1 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{-1+\sqrt{2}i}{3}\approx -0.333333333+0.471404521i$$

$$x=\frac{-\sqrt{2}i-1}{3}\approx -0.333333333-0.471404521i$$

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.

$$3x^{2}+2x+1=0$$

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

$$x=\frac{-2±\sqrt{2^{2}-4\times 3}}{2\times 3}$$

Pūrua 2.

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

Whakareatia -4 ki te 3.

$$x=\frac{-2±\sqrt{4-12}}{2\times 3}$$

Tāpiri 4 ki te -12.

$$x=\frac{-2±\sqrt{-8}}{2\times 3}$$

Tuhia te pūtakerua o te -8.

$$x=\frac{-2±2\sqrt{2}i}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-2±2\sqrt{2}i}{6}$$

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

$$x=\frac{-2+2\sqrt{2}i}{6}$$

Whakawehe -2+2i\sqrt{2} ki te 6.

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

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

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

Whakawehe -2-2i\sqrt{2} ki te 6.

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

Kua oti te whārite te whakatau.

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

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.

$$3x^{2}+2x+1=0$$

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

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

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

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

Whakawehea ngā taha e rua ki te 3.

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

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

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

Whakawehea te \frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{3}. Nā, tāpiria te pūrua o te \frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{1}{3}+\frac{1}{9}$$

Tāpiri -\frac{1}{3} ki te \frac{1}{9} 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.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{2}{9}$$

Tauwehea x^{2}+\frac{2}{3}x+\frac{1}{9}. 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}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{2}{9}$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{2}{9}}$$

Whakarūnātia.

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

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

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