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

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

Whakaoti mō x

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3}\approx 0.193712943$$

$$x=-\frac{\sqrt{10}}{6}-\frac{1}{3}\approx -0.86037961$$

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.

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

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

$$x=\frac{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\left(-\frac{1}{6}\right)}}{2}$$

Pūruatia \frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\left(-\frac{1}{6}\right)}}{2}$$

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

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+\frac{2}{3}}}{2}$$

Tāpiri \frac{4}{9} ki te \frac{2}{3} 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.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{10}{9}}}{2}$$

Tuhia te pūtakerua o te \frac{10}{9}.

$$x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2}$$

Nā, me whakaoti te whārite x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2} ina he tāpiri te ±. Tāpiri -\frac{2}{3} ki te \frac{\sqrt{10}}{3}.

$$x=\frac{\sqrt{10}-2}{2\times 3}$$

Whakawehe \frac{-2+\sqrt{10}}{3} ki te 2.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-\frac{2}{3}±\frac{\sqrt{10}}{3}}{2} ina he tango te ±. Tango \frac{\sqrt{10}}{3} mai i -\frac{2}{3}.

$$x=\frac{-\sqrt{10}-2}{2\times 3}$$

Whakawehe \frac{-2-\sqrt{10}}{3} ki te 2.

$$x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3} x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$

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.

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

Me tāpiri \frac{1}{6} ki ngā taha e rua o te whārite.

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

Mā te tango i te -\frac{1}{6} i a ia ake anō ka toe ko te 0.

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

Tango -\frac{1}{6} mai i 0.

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

Whakawehea te \frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{3}. Nā, tāpiria te pūrua o te \frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{6}+\frac{1}{9}$$

Tāpiri \frac{1}{6} ki te \frac{1}{9} 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.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{18}$$

Tauwehea x^{2}+\frac{2}{3}x+\frac{1}{9}. 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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{5}{18}$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{5}{18}}$$

Whakarūnātia.

$$x+\frac{1}{3}=\frac{\sqrt{10}}{6} x+\frac{1}{3}=-\frac{\sqrt{10}}{6}$$

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{10}}{6}-\frac{1}{3} x=-\frac{\sqrt{10}}{6}-\frac{1}{3}$$