Whakaoti i te 12x^2+3x=0

$$12 { x }^{ 2 } +3x=0$$

Whakaoti mō x

$$x=-\frac{1}{4}=-0.25$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(12x+3\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 12x+3=0.

$$x=0 x=-\frac{1}{4}$$

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.

$$12x^{2}+3x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 12 mō a, 3 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\times 12}$$

Tuhia te pūtakerua o te 3^{2}.

$$x=\frac{-3±3}{2\times 12}$$

Whakareatia 2 ki te 12.

$$x=\frac{-3±3}{24}$$

Nā, me whakaoti te whārite x=\frac{-3±3}{24} ina he tāpiri te ±. Tāpiri -3 ki te 3.

$$x=\frac{0}{24}$$

Whakawehe 0 ki te 24.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-3±3}{24} ina he tango te ±. Tango 3 mai i -3.

$$x=-\frac{6}{24}$$

Whakahekea te hautanga \frac{-6}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=-\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$x=0 x=-\frac{1}{4}$$

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.

$$12x^{2}+3x=0$$

Whakawehea ngā taha e rua ki te 12.

$$\frac{12x^{2}+3x}{12}=\frac{0}{12}$$

Mā te whakawehe ki te 12 ka wetekia te whakareanga ki te 12.

$$x^{2}+\frac{3}{12}x=\frac{0}{12}$$

Whakahekea te hautanga \frac{3}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}+\frac{1}{4}x=\frac{0}{12}$$

Whakawehe 0 ki te 12.

$$x^{2}+\frac{1}{4}x=0$$

Whakawehea te \frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{8}. Nā, tāpiria te pūrua o te \frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\left(\frac{1}{8}\right)^{2}$$

Pūruatia \frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Tauwehea x^{2}+\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{1}{64}$$

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

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Whakarūnātia.

$$x+\frac{1}{8}=\frac{1}{8} x+\frac{1}{8}=-\frac{1}{8}$$

Me tango \frac{1}{8} mai i ngā taha e rua o te whārite.

$$x=0 x=-\frac{1}{4}$$