Whakaoti i te 9x^2+6x=0

$$9 x ^ { 2 } + 6 x = 0$$

Whakaoti mō x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(9x+6\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 9x+6=0.

$$x=0 x=-\frac{2}{3}$$

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.

$$9x^{2}+6x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 6 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 9}$$

Tuhia te pūtakerua o te 6^{2}.

$$x=\frac{-6±6}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-6±6}{18}$$

Nā, me whakaoti te whārite x=\frac{-6±6}{18} ina he tāpiri te ±. Tāpiri -6 ki te 6.

$$x=\frac{0}{18}$$

Whakawehe 0 ki te 18.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-6±6}{18} ina he tango te ±. Tango 6 mai i -6.

$$x=-\frac{12}{18}$$

Whakahekea te hautanga \frac{-12}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Kua oti te whārite te whakatau.

$$x=0 x=-\frac{2}{3}$$

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.

$$9x^{2}+6x=0$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}+6x}{9}=\frac{0}{9}$$

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

$$x^{2}+\frac{6}{9}x=\frac{0}{9}$$

Whakahekea te hautanga \frac{6}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}+\frac{2}{3}x=\frac{0}{9}$$

Whakawehe 0 ki te 9.

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

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.

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

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

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

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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{1}{9}$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Whakarūnātia.

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

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

$$x=0 x=-\frac{2}{3}$$