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

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

Whakaoti mō x

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}\approx 0.457427108$$

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}\approx -1.457427108$$

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.

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

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

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 3\left(-2\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-3±\sqrt{9-12\left(-2\right)}}{2\times 3}$$

Whakareatia -12 ki te -2.

$$x=\frac{-3±\sqrt{9+24}}{2\times 3}$$

Tāpiri 9 ki te 24.

$$x=\frac{-3±\sqrt{33}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-3±\sqrt{33}}{6}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{33}}{6} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{6}$$

Whakawehe -3+\sqrt{33} ki te 6.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{33}}{6} ina he tango te ±. Tango \sqrt{33} mai i -3.

$$x=\frac{-\sqrt{33}-3}{6}$$

Whakawehe -3-\sqrt{33} ki te 6.

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2} x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

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.

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

Me tāpiri 2 ki ngā taha e rua o te whārite.

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

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

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

Tango -2 mai i 0.

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

Whakawehea ngā taha e rua ki te 3.

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

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

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

Whakawehe 3 ki te 3.

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

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.

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

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

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

Tāpiri \frac{2}{3} ki te \frac{1}{4} 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.

$$x^{2}+x+\frac{1}{4}=\frac{11}{12}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{11}{12}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{11}{12}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{33}}{6} x+\frac{1}{2}=-\frac{\sqrt{33}}{6}$$

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

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2} x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$