Whakaoti i te 5x^2-2x+1=0

$$5 x ^ { 2 } - 2 x + 1 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{1}{5}+\frac{2}{5}i=0.2+0.4i$$

$$x=\frac{1}{5}-\frac{2}{5}i=0.2-0.4i$$

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.

$$5x^{2}-2x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -2 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -2.

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

Whakareatia -4 ki te 5.

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

Tāpiri 4 ki te -20.

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

Tuhia te pūtakerua o te -16.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±4i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{2±4i}{10}$$

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

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

Whakawehe 2+4i ki te 10.

$$x=\frac{1}{5}+\frac{2}{5}i$$

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

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

Whakawehe 2-4i ki te 10.

$$x=\frac{1}{5}-\frac{2}{5}i$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{5}+\frac{2}{5}i x=\frac{1}{5}-\frac{2}{5}i$$

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.

$$5x^{2}-2x+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$5x^{2}-2x+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$5x^{2}-2x=-1$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-2x}{5}=-\frac{1}{5}$$

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

$$x^{2}-\frac{2}{5}x=-\frac{1}{5}$$

Whakawehea te -\frac{2}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{5}. Nā, tāpiria te pūrua o te -\frac{1}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{1}{5}+\left(-\frac{1}{5}\right)^{2}$$

Pūruatia -\frac{1}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{1}{5}+\frac{1}{25}$$

Tāpiri -\frac{1}{5} ki te \frac{1}{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.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{4}{25}$$

Tauwehea te x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{4}{25}$$

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

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}$$

Whakarūnātia.

$$x-\frac{1}{5}=\frac{2}{5}i x-\frac{1}{5}=-\frac{2}{5}i$$

Me tāpiri \frac{1}{5} ki ngā taha e rua o te whārite.

$$x=\frac{1}{5}+\frac{2}{5}i x=\frac{1}{5}-\frac{2}{5}i$$