Whakaoti i te 2x^2+8x+9=0

$$2 x ^ { 2 } + 8 x + 9 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{2}i}{2}-2\approx -2+0.707106781i$$

$$x=-\frac{\sqrt{2}i}{2}-2\approx -2-0.707106781i$$

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.

$$2x^{2}+8x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 8 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\times 9}}{2\times 2}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 2\times 9}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-8±\sqrt{64-8\times 9}}{2\times 2}$$

Whakareatia -8 ki te 9.

$$x=\frac{-8±\sqrt{64-72}}{2\times 2}$$

Tāpiri 64 ki te -72.

$$x=\frac{-8±\sqrt{-8}}{2\times 2}$$

Tuhia te pūtakerua o te -8.

$$x=\frac{-8±2\sqrt{2}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-8±2\sqrt{2}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-8±2\sqrt{2}i}{4} ina he tāpiri te ±. Tāpiri -8 ki te 2i\sqrt{2}.

$$x=\frac{-8+2\sqrt{2}i}{4}$$

Whakawehe -8+2i\sqrt{2} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{-8±2\sqrt{2}i}{4} ina he tango te ±. Tango 2i\sqrt{2} mai i -8.

$$x=\frac{-2\sqrt{2}i-8}{4}$$

Whakawehe -8-2i\sqrt{2} ki te 4.

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

Kua oti te whārite te whakatau.

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

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.

$$2x^{2}+8x+9=0$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$2x^{2}+8x+9-9=-9$$

Mā te tango i te 9 i a ia ake anō ka toe ko te 0.

$$2x^{2}+8x=-9$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+8x}{2}=-\frac{9}{2}$$

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

$$x^{2}+\frac{8}{2}x=-\frac{9}{2}$$

Whakawehe 8 ki te 2.

$$x^{2}+4x=-\frac{9}{2}$$

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.

$$x^{2}+4x+2^{2}=-\frac{9}{2}+2^{2}$$

Pūrua 2.

$$x^{2}+4x+4=-\frac{9}{2}+4$$

Tāpiri -\frac{9}{2} ki te 4.

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

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

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

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

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

Whakarūnātia.

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

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

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