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

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

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

Tangohia te 2x mai i ngā taha e rua.

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

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}.

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

Pūrua -2.

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

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12}}{2\times 3}$$

Tāpiri 4 ki te -12.

$$x=\frac{-\left(-2\right)±\sqrt{-8}}{2\times 3}$$

Tuhia te pūtakerua o te -8.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{2}i}{2\times 3}$$

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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 tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

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