Whakaoti i te 0=6x^2-3x+1

$$0 = 6 x ^ { 2 } - 3 x + 1$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{15}i}{12}+\frac{1}{4}\approx 0.25+0.322748612i$$

$$x=-\frac{\sqrt{15}i}{12}+\frac{1}{4}\approx 0.25-0.322748612i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$6x^{2}-3x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 6 mō a, -3 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6}}{2\times 6}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-24}}{2\times 6}$$

Tāpiri 9 ki te -24.

$$x=\frac{-\left(-3\right)±\sqrt{-15}}{2\times 6}$$

Tuhia te pūtakerua o te -15.

$$x=\frac{-\left(-3\right)±\sqrt{15}i}{2\times 6}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{15}i}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{3±\sqrt{15}i}{12}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{15}i}{12} ina he tāpiri te ±. Tāpiri 3 ki te i\sqrt{15}.

$$x=\frac{3+\sqrt{15}i}{12}$$

Whakawehe 3+i\sqrt{15} ki te 12.

$$x=\frac{\sqrt{15}i}{12}+\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{15}i}{12} ina he tango te ±. Tango i\sqrt{15} mai i 3.

$$x=\frac{-\sqrt{15}i+3}{12}$$

Whakawehe 3-i\sqrt{15} ki te 12.

$$x=-\frac{\sqrt{15}i}{12}+\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{15}i}{12}+\frac{1}{4} x=-\frac{\sqrt{15}i}{12}+\frac{1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$6x^{2}-3x+1=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$6x^{2}-3x=-1$$

Whakawehea ngā taha e rua ki te 6.

$$\frac{6x^{2}-3x}{6}=-\frac{1}{6}$$

Mā te whakawehe ki te 6 ka wetekia te whakareanga ki te 6.

$$x^{2}+\left(-\frac{3}{6}\right)x=-\frac{1}{6}$$

Whakahekea te hautanga \frac{-3}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{1}{2}x=-\frac{1}{6}$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{1}{6}+\left(-\frac{1}{4}\right)^{2}$$

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{6}+\frac{1}{16}$$

Tāpiri -\frac{1}{6} ki te \frac{1}{16} 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.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{5}{48}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{5}{48}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{5}{48}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{\sqrt{15}i}{12} x-\frac{1}{4}=-\frac{\sqrt{15}i}{12}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{15}i}{12}+\frac{1}{4} x=-\frac{\sqrt{15}i}{12}+\frac{1}{4}$$