Whakaoti i te 3x^2+9x+4=0

$$3 x ^ { 2 } + 9 x + 4 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{33}}{6}-\frac{3}{2}\approx -0.542572892$$

$$x=-\frac{\sqrt{33}}{6}-\frac{3}{2}\approx -2.457427108$$

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.

$$3x^{2}+9x+4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 9 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 3\times 4}}{2\times 3}$$

Pūrua 9.

$$x=\frac{-9±\sqrt{81-4\times 3\times 4}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-9±\sqrt{81-12\times 4}}{2\times 3}$$

Whakareatia -12 ki te 4.

$$x=\frac{-9±\sqrt{81-48}}{2\times 3}$$

Tāpiri 81 ki te -48.

$$x=\frac{-9±\sqrt{33}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-9±\sqrt{33}}{6}$$

Nā, me whakaoti te whārite x=\frac{-9±\sqrt{33}}{6} ina he tāpiri te ±. Tāpiri -9 ki te \sqrt{33}.

$$x=\frac{\sqrt{33}-9}{6}$$

Whakawehe -9+\sqrt{33} ki te 6.

$$x=\frac{\sqrt{33}}{6}-\frac{3}{2}$$

Nā, me whakaoti te whārite x=\frac{-9±\sqrt{33}}{6} ina he tango te ±. Tango \sqrt{33} mai i -9.

$$x=\frac{-\sqrt{33}-9}{6}$$

Whakawehe -9-\sqrt{33} ki te 6.

$$x=-\frac{\sqrt{33}}{6}-\frac{3}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{33}}{6}-\frac{3}{2} x=-\frac{\sqrt{33}}{6}-\frac{3}{2}$$

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.

$$3x^{2}+9x+4=0$$

Me tango 4 mai i ngā taha e rua o te whārite.

$$3x^{2}+9x+4-4=-4$$

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

$$3x^{2}+9x=-4$$

Whakawehea ngā taha e rua ki te 3.

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

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

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

Whakawehe 9 ki te 3.

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

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

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

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

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

Tāpiri -\frac{4}{3} ki te \frac{9}{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.

$$x^{2}+3x+\frac{9}{4}=\frac{11}{12}$$

Tauwehea te x^{2}+3x+\frac{9}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{11}{12}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{11}{12}}$$

Whakarūnātia.

$$x+\frac{3}{2}=\frac{\sqrt{33}}{6} x+\frac{3}{2}=-\frac{\sqrt{33}}{6}$$

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

$$x=\frac{\sqrt{33}}{6}-\frac{3}{2} x=-\frac{\sqrt{33}}{6}-\frac{3}{2}$$