Whakaoti i te 1+3x(1-x)=0

$$1+3x(1-x)=0$$

Whakaoti mō x

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2}\approx 1.263762616$$

$$x=-\frac{\sqrt{21}}{6}+\frac{1}{2}\approx -0.263762616$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te 1-x.

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

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.

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

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

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

Pūrua 3.

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

Whakareatia -4 ki te -3.

$$x=\frac{-3±\sqrt{9+12}}{2\left(-3\right)}$$

Tāpiri 9 ki te 12.

$$x=\frac{-3±\sqrt{21}}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-3±\sqrt{21}}{-6}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{21}}{-6} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{21}.

$$x=\frac{\sqrt{21}-3}{-6}$$

Whakawehe -3+\sqrt{21} ki te -6.

$$x=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{21}}{-6} ina he tango te ±. Tango \sqrt{21} mai i -3.

$$x=\frac{-\sqrt{21}-3}{-6}$$

Whakawehe -3-\sqrt{21} ki te -6.

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=-\frac{\sqrt{21}}{6}+\frac{1}{2} x=\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te 1-x.

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

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

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

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.

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

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe 3 ki te -3.

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

Whakawehe -1 ki te -3.

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

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.

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

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

$$x^{2}-x+\frac{1}{4}=\frac{1}{3}+\frac{1}{4}$$

Tāpiri \frac{1}{3} 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.

$$x^{2}-x+\frac{1}{4}=\frac{7}{12}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{7}{12}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{7}{12}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{21}}{6} x-\frac{1}{2}=-\frac{\sqrt{21}}{6}$$

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

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2} x=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$