Whakaoti i te x^2+4x+6=0

$${ x }^{ 2 } +4x+6=0$$

Whakaoti mō x (complex solution)

$$x=-2+\sqrt{2}i\approx -2+1.414213562i$$

$$x=-\sqrt{2}i-2\approx -2-1.414213562i$$

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.

$$x^{2}+4x+6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 4 mō b, me 6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 6}}{2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 6}}{2}$$

Whakareatia -4 ki te 6.

$$x=\frac{-4±\sqrt{16-24}}{2}$$

Tāpiri 16 ki te -24.

$$x=\frac{-4±\sqrt{-8}}{2}$$

Tuhia te pūtakerua o te -8.

$$x=\frac{-4±2\sqrt{2}i}{2}$$

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

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

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

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

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

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

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

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

Kua oti te whārite te whakatau.

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

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.

$$x^{2}+4x+6=0$$

Me tango 6 mai i ngā taha e rua o te whārite.

$$x^{2}+4x+6-6=-6$$

Mā te tango i te 6 i a ia ake anō ka toe ko te 0.

$$x^{2}+4x=-6$$

Whakawehea te 4, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 2. Nā, tāpiria te pūrua o te 2 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+4x+2^{2}=-6+2^{2}$$

Pūrua 2.

$$x^{2}+4x+4=-6+4$$

Tāpiri -6 ki te 4.

$$x^{2}+4x+4=-2$$

Tauwehea x^{2}+4x+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}.

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

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

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

Whakarūnātia.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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