Whakaoti i te 3k^2+11k+16=0

$$3 k ^ { 2 } + 11 k + 16 = 0$$

Whakaoti mō k

$$k=\frac{-11+\sqrt{71}i}{6}\approx -1.833333333+1.404358296i$$

$$k=\frac{-\sqrt{71}i-11}{6}\approx -1.833333333-1.404358296i$$

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.

$$3k^{2}+11k+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 11 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-11±\sqrt{11^{2}-4\times 3\times 16}}{2\times 3}$$

Pūrua 11.

$$k=\frac{-11±\sqrt{121-4\times 3\times 16}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$k=\frac{-11±\sqrt{121-12\times 16}}{2\times 3}$$

Whakareatia -12 ki te 16.

$$k=\frac{-11±\sqrt{121-192}}{2\times 3}$$

Tāpiri 121 ki te -192.

$$k=\frac{-11±\sqrt{-71}}{2\times 3}$$

Tuhia te pūtakerua o te -71.

$$k=\frac{-11±\sqrt{71}i}{2\times 3}$$

Whakareatia 2 ki te 3.

$$k=\frac{-11±\sqrt{71}i}{6}$$

Nā, me whakaoti te whārite k=\frac{-11±\sqrt{71}i}{6} ina he tāpiri te ±. Tāpiri -11 ki te i\sqrt{71}.

$$k=\frac{-11+\sqrt{71}i}{6}$$

Nā, me whakaoti te whārite k=\frac{-11±\sqrt{71}i}{6} ina he tango te ±. Tango i\sqrt{71} mai i -11.

$$k=\frac{-\sqrt{71}i-11}{6}$$

Kua oti te whārite te whakatau.

$$k=\frac{-11+\sqrt{71}i}{6} k=\frac{-\sqrt{71}i-11}{6}$$

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.

$$3k^{2}+11k+16=0$$

Me tango 16 mai i ngā taha e rua o te whārite.

$$3k^{2}+11k+16-16=-16$$

Mā te tango i te 16 i a ia ake anō ka toe ko te 0.

$$3k^{2}+11k=-16$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3k^{2}+11k}{3}=-\frac{16}{3}$$

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

$$k^{2}+\frac{11}{3}k=-\frac{16}{3}$$

Whakawehea te \frac{11}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{11}{6}. Nā, tāpiria te pūrua o te \frac{11}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$k^{2}+\frac{11}{3}k+\left(\frac{11}{6}\right)^{2}=-\frac{16}{3}+\left(\frac{11}{6}\right)^{2}$$

Pūruatia \frac{11}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$k^{2}+\frac{11}{3}k+\frac{121}{36}=-\frac{16}{3}+\frac{121}{36}$$

Tāpiri -\frac{16}{3} ki te \frac{121}{36} 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.

$$k^{2}+\frac{11}{3}k+\frac{121}{36}=-\frac{71}{36}$$

Tauwehea k^{2}+\frac{11}{3}k+\frac{121}{36}. 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}.

$$\left(k+\frac{11}{6}\right)^{2}=-\frac{71}{36}$$

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

$$\sqrt{\left(k+\frac{11}{6}\right)^{2}}=\sqrt{-\frac{71}{36}}$$

Whakarūnātia.

$$k+\frac{11}{6}=\frac{\sqrt{71}i}{6} k+\frac{11}{6}=-\frac{\sqrt{71}i}{6}$$

Me tango \frac{11}{6} mai i ngā taha e rua o te whārite.

$$k=\frac{-11+\sqrt{71}i}{6} k=\frac{-\sqrt{71}i-11}{6}$$