Whakaoti i te 5x^2+4x+3=0

$$5 x ^ { 2 } + 4 x + 3 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{-2+\sqrt{11}i}{5}\approx -0.4+0.663324958i$$

$$x=\frac{-\sqrt{11}i-2}{5}\approx -0.4-0.663324958i$$

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.

$$5x^{2}+4x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 4 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 5\times 3}}{2\times 5}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 5\times 3}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-4±\sqrt{16-20\times 3}}{2\times 5}$$

Whakareatia -20 ki te 3.

$$x=\frac{-4±\sqrt{16-60}}{2\times 5}$$

Tāpiri 16 ki te -60.

$$x=\frac{-4±\sqrt{-44}}{2\times 5}$$

Tuhia te pūtakerua o te -44.

$$x=\frac{-4±2\sqrt{11}i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-4±2\sqrt{11}i}{10}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{11}i}{10} ina he tāpiri te ±. Tāpiri -4 ki te 2i\sqrt{11}.

$$x=\frac{-4+2\sqrt{11}i}{10}$$

Whakawehe -4+2i\sqrt{11} ki te 10.

$$x=\frac{-2+\sqrt{11}i}{5}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{11}i}{10} ina he tango te ±. Tango 2i\sqrt{11} mai i -4.

$$x=\frac{-2\sqrt{11}i-4}{10}$$

Whakawehe -4-2i\sqrt{11} ki te 10.

$$x=\frac{-\sqrt{11}i-2}{5}$$

Kua oti te whārite te whakatau.

$$x=\frac{-2+\sqrt{11}i}{5} x=\frac{-\sqrt{11}i-2}{5}$$

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.

$$5x^{2}+4x+3=0$$

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

$$5x^{2}+4x+3-3=-3$$

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

$$5x^{2}+4x=-3$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+4x}{5}=-\frac{3}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}+\frac{4}{5}x=-\frac{3}{5}$$

Whakawehea te \frac{4}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{2}{5}. Nā, tāpiria te pūrua o te \frac{2}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=-\frac{3}{5}+\left(\frac{2}{5}\right)^{2}$$

Pūruatia \frac{2}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=-\frac{3}{5}+\frac{4}{25}$$

Tāpiri -\frac{3}{5} ki te \frac{4}{25} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=-\frac{11}{25}$$

Tauwehea x^{2}+\frac{4}{5}x+\frac{4}{25}. 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}.

$$\left(x+\frac{2}{5}\right)^{2}=-\frac{11}{25}$$

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

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{-\frac{11}{25}}$$

Whakarūnātia.

$$x+\frac{2}{5}=\frac{\sqrt{11}i}{5} x+\frac{2}{5}=-\frac{\sqrt{11}i}{5}$$

Me tango \frac{2}{5} mai i ngā taha e rua o te whārite.

$$x=\frac{-2+\sqrt{11}i}{5} x=\frac{-\sqrt{11}i-2}{5}$$