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

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

Whakaoti mō x

$$x=\frac{\sqrt{3}}{6}+\frac{1}{2}\approx 0.788675135$$

$$x=-\frac{\sqrt{3}}{6}+\frac{1}{2}\approx 0.211324865$$

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.

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

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

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

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 6}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-6\right)±\sqrt{36-24}}{2\times 6}$$

Tāpiri 36 ki te -24.

$$x=\frac{-\left(-6\right)±\sqrt{12}}{2\times 6}$$

Tuhia te pūtakerua o te 12.

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

Ko te tauaro o -6 ko 6.

$$x=\frac{6±2\sqrt{3}}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{6±2\sqrt{3}}{12}$$

Nā, me whakaoti te whārite x=\frac{6±2\sqrt{3}}{12} ina he tāpiri te ±. Tāpiri 6 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+6}{12}$$

Whakawehe 6+2\sqrt{3} ki te 12.

$$x=\frac{\sqrt{3}}{6}+\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{6±2\sqrt{3}}{12} ina he tango te ±. Tango 2\sqrt{3} mai i 6.

$$x=\frac{6-2\sqrt{3}}{12}$$

Whakawehe 6-2\sqrt{3} ki te 12.

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

Kua oti te whārite te whakatau.

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

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.

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

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

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

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

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

Whakawehea ngā taha e rua ki te 6.

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

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

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

Whakawehe -6 ki te 6.

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

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

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{6}+\frac{1}{4}$$

Tāpiri -\frac{1}{6} ki te \frac{1}{4} 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.

$$x^{2}-x+\frac{1}{4}=\frac{1}{12}$$

Tauwehea x^{2}-x+\frac{1}{4}. 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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{12}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{12}}$$

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

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