Whakaoti i te 3x^2+8x-1=0

$$3 { x }^{ 2 } +8x-1=0$$

Whakaoti mō x

$$x=\frac{\sqrt{19}-4}{3}\approx 0.119632981$$

$$x=\frac{-\sqrt{19}-4}{3}\approx -2.786299648$$

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.

$$3x^{2}+8x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 8 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 3\left(-1\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-8±\sqrt{64-12\left(-1\right)}}{2\times 3}$$

Whakareatia -12 ki te -1.

$$x=\frac{-8±\sqrt{64+12}}{2\times 3}$$

Tāpiri 64 ki te 12.

$$x=\frac{-8±\sqrt{76}}{2\times 3}$$

Tuhia te pūtakerua o te 76.

$$x=\frac{-8±2\sqrt{19}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-8±2\sqrt{19}}{6}$$

Nā, me whakaoti te whārite x=\frac{-8±2\sqrt{19}}{6} ina he tāpiri te ±. Tāpiri -8 ki te 2\sqrt{19}.

$$x=\frac{2\sqrt{19}-8}{6}$$

Whakawehe -8+2\sqrt{19} ki te 6.

$$x=\frac{\sqrt{19}-4}{3}$$

Nā, me whakaoti te whārite x=\frac{-8±2\sqrt{19}}{6} ina he tango te ±. Tango 2\sqrt{19} mai i -8.

$$x=\frac{-2\sqrt{19}-8}{6}$$

Whakawehe -8-2\sqrt{19} ki te 6.

$$x=\frac{-\sqrt{19}-4}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{19}-4}{3} x=\frac{-\sqrt{19}-4}{3}$$

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.

$$3x^{2}+8x-1=0$$

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

$$3x^{2}+8x-1-\left(-1\right)=-\left(-1\right)$$

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

$$3x^{2}+8x=-\left(-1\right)$$

Tango -1 mai i 0.

$$3x^{2}+8x=1$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+8x}{3}=\frac{1}{3}$$

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

$$x^{2}+\frac{8}{3}x=\frac{1}{3}$$

Whakawehea te \frac{8}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{4}{3}. Nā, tāpiria te pūrua o te \frac{4}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\frac{1}{3}+\left(\frac{4}{3}\right)^{2}$$

Pūruatia \frac{4}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{1}{3}+\frac{16}{9}$$

Tāpiri \frac{1}{3} ki te \frac{16}{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.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{19}{9}$$

Tauwehea x^{2}+\frac{8}{3}x+\frac{16}{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}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{19}{9}$$

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

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{19}{9}}$$

Whakarūnātia.

$$x+\frac{4}{3}=\frac{\sqrt{19}}{3} x+\frac{4}{3}=-\frac{\sqrt{19}}{3}$$

Me tango \frac{4}{3} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{19}-4}{3} x=\frac{-\sqrt{19}-4}{3}$$