Whakaoti i te -3x^2-4x+2=0

$$- 3 x ^ { 2 } - 4 x + 2 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{10}-2}{3}\approx 0.387425887$$

$$x=\frac{-\sqrt{10}-2}{3}\approx -1.72075922$$

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.

$$-3x^{2}-4x+2=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12\times 2}}{2\left(-3\right)}$$

Whakareatia 12 ki te 2.

$$x=\frac{-\left(-4\right)±\sqrt{16+24}}{2\left(-3\right)}$$

Tāpiri 16 ki te 24.

$$x=\frac{-\left(-4\right)±\sqrt{40}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-\left(-4\right)±2\sqrt{10}}{2\left(-3\right)}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{10}}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{4±2\sqrt{10}}{-6}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{10}}{-6} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+4}{-6}$$

Whakawehe 4+2\sqrt{10} ki te -6.

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

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{10}}{-6} ina he tango te ±. Tango 2\sqrt{10} mai i 4.

$$x=\frac{4-2\sqrt{10}}{-6}$$

Whakawehe 4-2\sqrt{10} ki te -6.

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

Kua oti te whārite te whakatau.

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

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.

$$-3x^{2}-4x+2=0$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$-3x^{2}-4x+2-2=-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$-3x^{2}-4x=-2$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3x^{2}-4x}{-3}=-\frac{2}{-3}$$

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

$$x^{2}+\left(-\frac{4}{-3}\right)x=-\frac{2}{-3}$$

Whakawehe -4 ki te -3.

$$x^{2}+\frac{4}{3}x=-\frac{2}{-3}$$

Whakawehe -2 ki te -3.

$$x^{2}+\frac{4}{3}x=\frac{2}{3}$$

Whakawehea te \frac{4}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{2}{3}. Nā, tāpiria te pūrua o te \frac{2}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{2}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$

Tāpiri \frac{2}{3} ki te \frac{4}{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.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{10}{9}$$

Tauwehea x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{10}{9}$$

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

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Whakarūnātia.

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

Me tango \frac{2}{3} mai i ngā taha e rua o te whārite.

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