Whakaoti i te 5x^2=2x-2

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

Whakaoti mō x (complex solution)

$$x=\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

$$x=\frac{1}{5}-\frac{3}{5}i=0.2-0.6i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 2x mai i ngā taha e rua.

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

Me tāpiri te 2 ki ngā taha e rua.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te 5.

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

Whakareatia -20 ki te 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-40}}{2\times 5}$$

Tāpiri 4 ki te -40.

$$x=\frac{-\left(-2\right)±\sqrt{-36}}{2\times 5}$$

Tuhia te pūtakerua o te -36.

$$x=\frac{-\left(-2\right)±6i}{2\times 5}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±6i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{2±6i}{10}$$

Nā, me whakaoti te whārite x=\frac{2±6i}{10} ina he tāpiri te ±. Tāpiri 2 ki te 6i.

$$x=\frac{2+6i}{10}$$

Whakawehe 2+6i ki te 10.

$$x=\frac{1}{5}+\frac{3}{5}i$$

Nā, me whakaoti te whārite x=\frac{2±6i}{10} ina he tango te ±. Tango 6i mai i 2.

$$x=\frac{2-6i}{10}$$

Whakawehe 2-6i ki te 10.

$$x=\frac{1}{5}-\frac{3}{5}i$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{5}+\frac{3}{5}i x=\frac{1}{5}-\frac{3}{5}i$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 2x mai i ngā taha e rua.

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

Whakawehea ngā taha e rua ki te 5.

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

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

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

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.

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

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

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

Tāpiri -\frac{2}{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.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{9}{25}$$

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

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{9}{25}$$

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

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{9}{25}}$$

Whakarūnātia.

$$x-\frac{1}{5}=\frac{3}{5}i x-\frac{1}{5}=-\frac{3}{5}i$$

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

$$x=\frac{1}{5}+\frac{3}{5}i x=\frac{1}{5}-\frac{3}{5}i$$