Whakaoti i te 5x^2-4x+10=0

$$5 { x }^{ 2 } -4x+10 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{2+\sqrt{46}i}{5}\approx 0.4+1.356465997i$$

$$x=\frac{-\sqrt{46}i+2}{5}\approx 0.4-1.356465997i$$

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.

$$5x^{2}-4x+10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -4 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\times 10}}{2\times 5}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\times 10}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\times 10}}{2\times 5}$$

Whakareatia -20 ki te 10.

$$x=\frac{-\left(-4\right)±\sqrt{16-200}}{2\times 5}$$

Tāpiri 16 ki te -200.

$$x=\frac{-\left(-4\right)±\sqrt{-184}}{2\times 5}$$

Tuhia te pūtakerua o te -184.

$$x=\frac{-\left(-4\right)±2\sqrt{46}i}{2\times 5}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{46}i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{4±2\sqrt{46}i}{10}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{46}i}{10} ina he tāpiri te ±. Tāpiri 4 ki te 2i\sqrt{46}.

$$x=\frac{4+2\sqrt{46}i}{10}$$

Whakawehe 4+2i\sqrt{46} ki te 10.

$$x=\frac{2+\sqrt{46}i}{5}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{46}i}{10} ina he tango te ±. Tango 2i\sqrt{46} mai i 4.

$$x=\frac{-2\sqrt{46}i+4}{10}$$

Whakawehe 4-2i\sqrt{46} ki te 10.

$$x=\frac{-\sqrt{46}i+2}{5}$$

Kua oti te whārite te whakatau.

$$x=\frac{2+\sqrt{46}i}{5} x=\frac{-\sqrt{46}i+2}{5}$$

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.

$$5x^{2}-4x+10=0$$

Me tango 10 mai i ngā taha e rua o te whārite.

$$5x^{2}-4x+10-10=-10$$

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

$$5x^{2}-4x=-10$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-4x}{5}=-\frac{10}{5}$$

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

$$x^{2}-\frac{4}{5}x=-\frac{10}{5}$$

Whakawehe -10 ki te 5.

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

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.

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

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

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

Tāpiri -2 ki te \frac{4}{25}.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{46}{25}$$

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

$$\left(x-\frac{2}{5}\right)^{2}=-\frac{46}{25}$$

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

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{-\frac{46}{25}}$$

Whakarūnātia.

$$x-\frac{2}{5}=\frac{\sqrt{46}i}{5} x-\frac{2}{5}=-\frac{\sqrt{46}i}{5}$$

Me tāpiri \frac{2}{5} ki ngā taha e rua o te whārite.

$$x=\frac{2+\sqrt{46}i}{5} x=\frac{-\sqrt{46}i+2}{5}$$