Whakaoti i te 3x^2+3x+5=0

$$3 { x }^{ 2 } +3x+5=0$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{51}i}{6}-\frac{1}{2}\approx -0.5+1.190238071i$$

$$x=-\frac{\sqrt{51}i}{6}-\frac{1}{2}\approx -0.5-1.190238071i$$

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.

$$3x^{2}+3x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 3 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\times 5}}{2\times 3}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 3\times 5}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-3±\sqrt{9-12\times 5}}{2\times 3}$$

Whakareatia -12 ki te 5.

$$x=\frac{-3±\sqrt{9-60}}{2\times 3}$$

Tāpiri 9 ki te -60.

$$x=\frac{-3±\sqrt{-51}}{2\times 3}$$

Tuhia te pūtakerua o te -51.

$$x=\frac{-3±\sqrt{51}i}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-3±\sqrt{51}i}{6}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{51}i}{6} ina he tāpiri te ±. Tāpiri -3 ki te i\sqrt{51}.

$$x=\frac{-3+\sqrt{51}i}{6}$$

Whakawehe -3+i\sqrt{51} ki te 6.

$$x=\frac{\sqrt{51}i}{6}-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{51}i}{6} ina he tango te ±. Tango i\sqrt{51} mai i -3.

$$x=\frac{-\sqrt{51}i-3}{6}$$

Whakawehe -3-i\sqrt{51} ki te 6.

$$x=-\frac{\sqrt{51}i}{6}-\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{51}i}{6}-\frac{1}{2} x=-\frac{\sqrt{51}i}{6}-\frac{1}{2}$$

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.

$$3x^{2}+3x+5=0$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$3x^{2}+3x+5-5=-5$$

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

$$3x^{2}+3x=-5$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+3x}{3}=-\frac{5}{3}$$

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

$$x^{2}+\frac{3}{3}x=-\frac{5}{3}$$

Whakawehe 3 ki te 3.

$$x^{2}+x=-\frac{5}{3}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{5}{3}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=-\frac{5}{3}+\frac{1}{4}$$

Tāpiri -\frac{5}{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.

$$x^{2}+x+\frac{1}{4}=-\frac{17}{12}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{17}{12}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{17}{12}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{51}i}{6} x+\frac{1}{2}=-\frac{\sqrt{51}i}{6}$$

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

$$x=\frac{\sqrt{51}i}{6}-\frac{1}{2} x=-\frac{\sqrt{51}i}{6}-\frac{1}{2}$$