Whakaoti i te 3y^2+3y=0

$$3 y ^ { 2 } + 3 y = 0$$

Whakaoti mō y

$$y=-1$$

$$y=0$$

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.

$$3y^{2}+3y=0$$

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

$$y=\frac{-3±\sqrt{3^{2}}}{2\times 3}$$

Tuhia te pūtakerua o te 3^{2}.

$$y=\frac{-3±3}{2\times 3}$$

Whakareatia 2 ki te 3.

$$y=\frac{-3±3}{6}$$

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

$$y=\frac{0}{6}$$

Whakawehe 0 ki te 6.

$$y=0$$

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

$$y=-\frac{6}{6}$$

Whakawehe -6 ki te 6.

$$y=-1$$

Kua oti te whārite te whakatau.

$$y=0 y=-1$$

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.

$$3y^{2}+3y=0$$

Whakawehea ngā taha e rua ki te 3.

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

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

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

Whakawehe 3 ki te 3.

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

Whakawehe 0 ki te 3.

$$y^{2}+y=0$$

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.

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

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

$$y^{2}+y+\frac{1}{4}=\frac{1}{4}$$

Tauwehea y^{2}+y+\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}.

$$\left(y+\frac{1}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(y+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$y+\frac{1}{2}=\frac{1}{2} y+\frac{1}{2}=-\frac{1}{2}$$

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

$$y=0 y=-1$$