Whakaoti i te 9x^2+9x=1

$$9 x ^ { 2 } + 9 x = 1$$

Whakaoti mō x

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2}\approx 0.100925213$$

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{2}\approx -1.100925213$$

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.

$$9x^{2}+9x=1$$

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

$$9x^{2}+9x-1=1-1$$

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

$$9x^{2}+9x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 9 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 9.

$$x=\frac{-9±\sqrt{81-4\times 9\left(-1\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-9±\sqrt{81-36\left(-1\right)}}{2\times 9}$$

Whakareatia -36 ki te -1.

$$x=\frac{-9±\sqrt{81+36}}{2\times 9}$$

Tāpiri 81 ki te 36.

$$x=\frac{-9±\sqrt{117}}{2\times 9}$$

Tuhia te pūtakerua o te 117.

$$x=\frac{-9±3\sqrt{13}}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-9±3\sqrt{13}}{18}$$

Nā, me whakaoti te whārite x=\frac{-9±3\sqrt{13}}{18} ina he tāpiri te ±. Tāpiri -9 ki te 3\sqrt{13}.

$$x=\frac{3\sqrt{13}-9}{18}$$

Whakawehe -9+3\sqrt{13} ki te 18.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-9±3\sqrt{13}}{18} ina he tango te ±. Tango 3\sqrt{13} mai i -9.

$$x=\frac{-3\sqrt{13}-9}{18}$$

Whakawehe -9-3\sqrt{13} ki te 18.

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2} x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$

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.

$$9x^{2}+9x=1$$

Whakawehea ngā taha e rua ki te 9.

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

Mā te whakawehe ki te 9 ka wetekia te whakareanga ki te 9.

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

Whakawehe 9 ki te 9.

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

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{1}{9}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{1}{9}+\frac{1}{4}$$

Tāpiri \frac{1}{9} 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.

$$x^{2}+x+\frac{1}{4}=\frac{13}{36}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{13}{36}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{13}}{6} x+\frac{1}{2}=-\frac{\sqrt{13}}{6}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2} x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$