Whakaoti i te 3x^2=5x-10

$$3 { x }^{ 2 } =5x-10$$

Whakaoti mō x (complex solution)

$$x=\frac{5+\sqrt{95}i}{6}\approx 0.833333333+1.624465724i$$

$$x=\frac{-\sqrt{95}i+5}{6}\approx 0.833333333-1.624465724i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 5x mai i ngā taha e rua.

$$3x^{2}-5x=-10$$

Me tāpiri te 10 ki ngā taha e rua.

$$3x^{2}-5x+10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -5 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times 10}}{2\times 3}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times 10}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12\times 10}}{2\times 3}$$

Whakareatia -12 ki te 10.

$$x=\frac{-\left(-5\right)±\sqrt{25-120}}{2\times 3}$$

Tāpiri 25 ki te -120.

$$x=\frac{-\left(-5\right)±\sqrt{-95}}{2\times 3}$$

Tuhia te pūtakerua o te -95.

$$x=\frac{-\left(-5\right)±\sqrt{95}i}{2\times 3}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{95}i}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{5±\sqrt{95}i}{6}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{95}i}{6} ina he tāpiri te ±. Tāpiri 5 ki te i\sqrt{95}.

$$x=\frac{5+\sqrt{95}i}{6}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{95}i}{6} ina he tango te ±. Tango i\sqrt{95} mai i 5.

$$x=\frac{-\sqrt{95}i+5}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{5+\sqrt{95}i}{6} x=\frac{-\sqrt{95}i+5}{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 5x mai i ngā taha e rua.

$$3x^{2}-5x=-10$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}-5x}{3}=-\frac{10}{3}$$

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

$$x^{2}-\frac{5}{3}x=-\frac{10}{3}$$

Whakawehea te -\frac{5}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{6}. Nā, tāpiria te pūrua o te -\frac{5}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{5}{6}\right)^{2}$$

Pūruatia -\frac{5}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{10}{3}+\frac{25}{36}$$

Tāpiri -\frac{10}{3} ki te \frac{25}{36} 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.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{95}{36}$$

Tauwehea x^{2}-\frac{5}{3}x+\frac{25}{36}. 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}.

$$\left(x-\frac{5}{6}\right)^{2}=-\frac{95}{36}$$

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

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{95}{36}}$$

Whakarūnātia.

$$x-\frac{5}{6}=\frac{\sqrt{95}i}{6} x-\frac{5}{6}=-\frac{\sqrt{95}i}{6}$$

Me tāpiri \frac{5}{6} ki ngā taha e rua o te whārite.

$$x=\frac{5+\sqrt{95}i}{6} x=\frac{-\sqrt{95}i+5}{6}$$