Whakaoti i te 2x^2+12x+40=0

$$2x^2+12x+40=0$$

Whakaoti mō x (complex solution)

$$x=-3+\sqrt{11}i\approx -3+3.31662479i$$

$$x=-\sqrt{11}i-3\approx -3-3.31662479i$$

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.

$$2x^{2}+12x+40=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 12 mō b, me 40 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\times 40}}{2\times 2}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 2\times 40}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-12±\sqrt{144-8\times 40}}{2\times 2}$$

Whakareatia -8 ki te 40.

$$x=\frac{-12±\sqrt{144-320}}{2\times 2}$$

Tāpiri 144 ki te -320.

$$x=\frac{-12±\sqrt{-176}}{2\times 2}$$

Tuhia te pūtakerua o te -176.

$$x=\frac{-12±4\sqrt{11}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-12±4\sqrt{11}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-12±4\sqrt{11}i}{4} ina he tāpiri te ±. Tāpiri -12 ki te 4i\sqrt{11}.

$$x=\frac{-12+4\sqrt{11}i}{4}$$

Whakawehe -12+4i\sqrt{11} ki te 4.

$$x=-3+\sqrt{11}i$$

Nā, me whakaoti te whārite x=\frac{-12±4\sqrt{11}i}{4} ina he tango te ±. Tango 4i\sqrt{11} mai i -12.

$$x=\frac{-4\sqrt{11}i-12}{4}$$

Whakawehe -12-4i\sqrt{11} ki te 4.

$$x=-\sqrt{11}i-3$$

Kua oti te whārite te whakatau.

$$x=-3+\sqrt{11}i x=-\sqrt{11}i-3$$

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.

$$2x^{2}+12x+40=0$$

Me tango 40 mai i ngā taha e rua o te whārite.

$$2x^{2}+12x+40-40=-40$$

Mā te tango i te 40 i a ia ake anō ka toe ko te 0.

$$2x^{2}+12x=-40$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+12x}{2}=-\frac{40}{2}$$

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

$$x^{2}+\frac{12}{2}x=-\frac{40}{2}$$

Whakawehe 12 ki te 2.

$$x^{2}+6x=-\frac{40}{2}$$

Whakawehe -40 ki te 2.

$$x^{2}+6x=-20$$

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

$$x^{2}+6x+3^{2}=-20+3^{2}$$

Pūrua 3.

$$x^{2}+6x+9=-20+9$$

Tāpiri -20 ki te 9.

$$x^{2}+6x+9=-11$$

Tauwehea te x^{2}+6x+9. 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}.

$$\left(x+3\right)^{2}=-11$$

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

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{-11}$$

Whakarūnātia.

$$x+3=\sqrt{11}i x+3=-\sqrt{11}i$$

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

$$x=-3+\sqrt{11}i x=-\sqrt{11}i-3$$