Whakaoti i te x-x^2=3

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

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{11}i+1}{2}\approx 0.5-1.658312395i$$

$$x=\frac{1+\sqrt{11}i}{2}\approx 0.5+1.658312395i$$

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.

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

Me tango 3 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te -3.

$$x=\frac{-1±\sqrt{1-12}}{2\left(-1\right)}$$

Tāpiri 1 ki te -12.

$$x=\frac{-1±\sqrt{-11}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te -11.

$$x=\frac{-1±\sqrt{11}i}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-1±\sqrt{11}i}{-2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{11}i}{-2} ina he tāpiri te ±. Tāpiri -1 ki te i\sqrt{11}.

$$x=\frac{-1+\sqrt{11}i}{-2}$$

Whakawehe -1+i\sqrt{11} ki te -2.

$$x=\frac{-\sqrt{11}i+1}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{11}i}{-2} ina he tango te ±. Tango i\sqrt{11} mai i -1.

$$x=\frac{-\sqrt{11}i-1}{-2}$$

Whakawehe -1-i\sqrt{11} ki te -2.

$$x=\frac{1+\sqrt{11}i}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

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.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe 1 ki te -1.

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

Whakawehe 3 ki te -1.

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

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.

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

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

$$x^{2}-x+\frac{1}{4}=-3+\frac{1}{4}$$

Tāpiri -3 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{11}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{11}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{11}i}{2} x-\frac{1}{2}=-\frac{\sqrt{11}i}{2}$$

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

$$x=\frac{1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+1}{2}$$