Whakaoti i te 2x-2x^2=12

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

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{23}i}{2}\approx 0.5+2.397915762i$$

$$x=\frac{-\sqrt{23}i+1}{2}\approx 0.5-2.397915762i$$

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.

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

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

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

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

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -2.

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

Whakareatia 8 ki te -12.

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

Tāpiri 4 ki te -96.

$$x=\frac{-2±\sqrt{-92}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te -92.

$$x=\frac{-2±2\sqrt{23}i}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-2±2\sqrt{23}i}{-4}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{23}i}{-4} ina he tāpiri te ±. Tāpiri -2 ki te 2i\sqrt{23}.

$$x=\frac{-2+2\sqrt{23}i}{-4}$$

Whakawehe -2+2i\sqrt{23} ki te -4.

$$x=\frac{-\sqrt{23}i+1}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{23}i}{-4} ina he tango te ±. Tango 2i\sqrt{23} mai i -2.

$$x=\frac{-2\sqrt{23}i-2}{-4}$$

Whakawehe -2-2i\sqrt{23} ki te -4.

$$x=\frac{1+\sqrt{23}i}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{23}i+1}{2} x=\frac{1+\sqrt{23}i}{2}$$

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.

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

Whakawehea ngā taha e rua ki te -2.

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

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

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

Whakawehe 2 ki te -2.

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

Whakawehe 12 ki te -2.

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

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-6+\left(-\frac{1}{2}\right)^{2}$$

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=-6+\frac{1}{4}$$

Tāpiri -6 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{23}{4}$$

Tauwehea te x^{2}-x+\frac{1}{4}. 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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{23}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{23}i}{2} x-\frac{1}{2}=-\frac{\sqrt{23}i}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=\frac{1+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+1}{2}$$