Whakaoti i te (x^2+1)+2x=-10

$$( { x }^{ 2 } +1)+2x=-10$$

Whakaoti mō x (complex solution)

$$x=-1+\sqrt{10}i\approx -1+3.16227766i$$

$$x=-\sqrt{10}i-1\approx -1-3.16227766i$$

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.

$$x^{2}+2x+1=-10$$

Me tāpiri 10 ki ngā taha e rua o te whārite.

$$x^{2}+2x+1-\left(-10\right)=-10-\left(-10\right)$$

Mā te tango i te -10 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x+1-\left(-10\right)=0$$

Tango -10 mai i 1.

$$x^{2}+2x+11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 2 mō b, me 11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 11}}{2}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\times 11}}{2}$$

Whakareatia -4 ki te 11.

$$x=\frac{-2±\sqrt{4-44}}{2}$$

Tāpiri 4 ki te -44.

$$x=\frac{-2±\sqrt{-40}}{2}$$

Tuhia te pūtakerua o te -40.

$$x=\frac{-2±2\sqrt{10}i}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}i}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2i\sqrt{10}.

$$x=\frac{-2+2\sqrt{10}i}{2}$$

Whakawehe -2+2i\sqrt{10} ki te 2.

$$x=-1+\sqrt{10}i$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}i}{2} ina he tango te ±. Tango 2i\sqrt{10} mai i -2.

$$x=\frac{-2\sqrt{10}i-2}{2}$$

Whakawehe -2-2i\sqrt{10} ki te 2.

$$x=-\sqrt{10}i-1$$

Kua oti te whārite te whakatau.

$$x=-1+\sqrt{10}i x=-\sqrt{10}i-1$$

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.

$$x^{2}+2x+1=-10$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=-10$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{-10}$$

Whakarūnātia.

$$x+1=\sqrt{10}i x+1=-\sqrt{10}i$$

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

$$x=-1+\sqrt{10}i x=-\sqrt{10}i-1$$