Whakaoti i te 3x^2+4x-1=0

$$3 x ^ { 2 } + 4 x - 1 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{7}-2}{3}\approx 0.215250437$$

$$x=\frac{-\sqrt{7}-2}{3}\approx -1.54858377$$

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.

$$3x^{2}+4x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 4 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-1\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12\left(-1\right)}}{2\times 3}$$

Whakareatia -12 ki te -1.

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

Tāpiri 16 ki te 12.

$$x=\frac{-4±\sqrt{28}}{2\times 3}$$

Tuhia te pūtakerua o te 28.

$$x=\frac{-4±2\sqrt{7}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-4±2\sqrt{7}}{6}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{7}}{6} ina he tāpiri te ±. Tāpiri -4 ki te 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-4}{6}$$

Whakawehe -4+2\sqrt{7} ki te 6.

$$x=\frac{\sqrt{7}-2}{3}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{7}}{6} ina he tango te ±. Tango 2\sqrt{7} mai i -4.

$$x=\frac{-2\sqrt{7}-4}{6}$$

Whakawehe -4-2\sqrt{7} ki te 6.

$$x=\frac{-\sqrt{7}-2}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$

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.

$$3x^{2}+4x-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$3x^{2}+4x-1-\left(-1\right)=-\left(-1\right)$$

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

$$3x^{2}+4x=-\left(-1\right)$$

Tango -1 mai i 0.

$$3x^{2}+4x=1$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+4x}{3}=\frac{1}{3}$$

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

$$x^{2}+\frac{4}{3}x=\frac{1}{3}$$

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.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{1}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{3}+\frac{4}{9}$$

Tāpiri \frac{1}{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.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{9}$$

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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{7}{9}$$

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

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Whakarūnātia.

$$x+\frac{2}{3}=\frac{\sqrt{7}}{3} x+\frac{2}{3}=-\frac{\sqrt{7}}{3}$$

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

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$