Whakaoti i te 3q^2+q-14=0

$$3 q ^ { 2 } + q - 14 = 0$$

Whakaoti mō q

$$q = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$q=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3q^{2}+aq+bq-14. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=3\left(-14\right)=-42$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -42.

$$-1,42 -2,21 -3,14 -6,7$$

Tātaihia te tapeke mō ia takirua.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-6 b=7$$

Tuhia anō te 3q^{2}+q-14 hei \left(3q^{2}-6q\right)+\left(7q-14\right).

$$\left(3q^{2}-6q\right)+\left(7q-14\right)$$

Tauwehea te 3q i te tuatahi me te 7 i te rōpū tuarua.

$$3q\left(q-2\right)+7\left(q-2\right)$$

Whakatauwehea atu te kīanga pātahi q-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(q-2\right)\left(3q+7\right)$$

Hei kimi otinga whārite, me whakaoti te q-2=0 me te 3q+7=0.

$$q=2 q=-\frac{7}{3}$$

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.

$$3q^{2}+q-14=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 1 mō b, me -14 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-1±\sqrt{1^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Pūrua 1.

$$q=\frac{-1±\sqrt{1-4\times 3\left(-14\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$q=\frac{-1±\sqrt{1-12\left(-14\right)}}{2\times 3}$$

Whakareatia -12 ki te -14.

$$q=\frac{-1±\sqrt{1+168}}{2\times 3}$$

Tāpiri 1 ki te 168.

$$q=\frac{-1±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$q=\frac{-1±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$q=\frac{-1±13}{6}$$

Nā, me whakaoti te whārite q=\frac{-1±13}{6} ina he tāpiri te ±. Tāpiri -1 ki te 13.

$$q=\frac{12}{6}$$

Whakawehe 12 ki te 6.

$$q=2$$

Nā, me whakaoti te whārite q=\frac{-1±13}{6} ina he tango te ±. Tango 13 mai i -1.

$$q=-\frac{14}{6}$$

Whakahekea te hautanga \frac{-14}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$q=-\frac{7}{3}$$

Kua oti te whārite te whakatau.

$$q=2 q=-\frac{7}{3}$$

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.

$$3q^{2}+q-14=0$$

Me tāpiri 14 ki ngā taha e rua o te whārite.

$$3q^{2}+q-14-\left(-14\right)=-\left(-14\right)$$

Mā te tango i te -14 i a ia ake anō ka toe ko te 0.

$$3q^{2}+q=-\left(-14\right)$$

Tango -14 mai i 0.

$$3q^{2}+q=14$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3q^{2}+q}{3}=\frac{14}{3}$$

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

$$q^{2}+\frac{1}{3}q=\frac{14}{3}$$

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$q^{2}+\frac{1}{3}q+\left(\frac{1}{6}\right)^{2}=\frac{14}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$q^{2}+\frac{1}{3}q+\frac{1}{36}=\frac{14}{3}+\frac{1}{36}$$

Tāpiri \frac{14}{3} ki te \frac{1}{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.

$$q^{2}+\frac{1}{3}q+\frac{1}{36}=\frac{169}{36}$$

Tauwehea q^{2}+\frac{1}{3}q+\frac{1}{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}.

$$\left(q+\frac{1}{6}\right)^{2}=\frac{169}{36}$$

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

$$\sqrt{\left(q+\frac{1}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Whakarūnātia.

$$q+\frac{1}{6}=\frac{13}{6} q+\frac{1}{6}=-\frac{13}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

$$q=2 q=-\frac{7}{3}$$