Whakaoti i te 3q^2-19q+16=0

$$3 q ^ { 2 } - 19 q + 16 = 0$$

Whakaoti mō q

$$q=1$$

$$q = \frac{16}{3} = 5\frac{1}{3} \approx 5.333333333$$

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+16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-19 ab=3\times 16=48$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Tātaihia te tapeke mō ia takirua.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

Ko te otinga te takirua ka hoatu i te tapeke -19.

$$a=-16 b=-3$$

Tuhia anō te 3q^{2}-19q+16 hei \left(3q^{2}-16q\right)+\left(-3q+16\right).

$$\left(3q^{2}-16q\right)+\left(-3q+16\right)$$

Tauwehea te q i te tuatahi me te -1 i te rōpū tuarua.

$$q\left(3q-16\right)-\left(3q-16\right)$$

Whakatauwehea atu te kīanga pātahi 3q-16 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3q-16\right)\left(q-1\right)$$

Hei kimi otinga whārite, me whakaoti te 3q-16=0 me te q-1=0.

$$q=\frac{16}{3} q=1$$

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.

$$3q^{2}-19q+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -19 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 3\times 16}}{2\times 3}$$

Pūrua -19.

$$q=\frac{-\left(-19\right)±\sqrt{361-4\times 3\times 16}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$q=\frac{-\left(-19\right)±\sqrt{361-12\times 16}}{2\times 3}$$

Whakareatia -12 ki te 16.

$$q=\frac{-\left(-19\right)±\sqrt{361-192}}{2\times 3}$$

Tāpiri 361 ki te -192.

$$q=\frac{-\left(-19\right)±\sqrt{169}}{2\times 3}$$

Tuhia te pūtakerua o te 169.

$$q=\frac{-\left(-19\right)±13}{2\times 3}$$

Ko te tauaro o -19 ko 19.

$$q=\frac{19±13}{2\times 3}$$

Whakareatia 2 ki te 3.

$$q=\frac{19±13}{6}$$

Nā, me whakaoti te whārite q=\frac{19±13}{6} ina he tāpiri te ±. Tāpiri 19 ki te 13.

$$q=\frac{32}{6}$$

Whakahekea te hautanga \frac{32}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$q=\frac{16}{3}$$

Nā, me whakaoti te whārite q=\frac{19±13}{6} ina he tango te ±. Tango 13 mai i 19.

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

Whakawehe 6 ki te 6.

$$q=1$$

Kua oti te whārite te whakatau.

$$q=\frac{16}{3} q=1$$

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.

$$3q^{2}-19q+16=0$$

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

$$3q^{2}-19q+16-16=-16$$

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

$$3q^{2}-19q=-16$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3q^{2}-19q}{3}=-\frac{16}{3}$$

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

$$q^{2}-\frac{19}{3}q=-\frac{16}{3}$$

Whakawehea te -\frac{19}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{19}{6}. Nā, tāpiria te pūrua o te -\frac{19}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$q^{2}-\frac{19}{3}q+\left(-\frac{19}{6}\right)^{2}=-\frac{16}{3}+\left(-\frac{19}{6}\right)^{2}$$

Pūruatia -\frac{19}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$q^{2}-\frac{19}{3}q+\frac{361}{36}=-\frac{16}{3}+\frac{361}{36}$$

Tāpiri -\frac{16}{3} ki te \frac{361}{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.

$$q^{2}-\frac{19}{3}q+\frac{361}{36}=\frac{169}{36}$$

Tauwehea te q^{2}-\frac{19}{3}q+\frac{361}{36}. 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}.

$$\left(q-\frac{19}{6}\right)^{2}=\frac{169}{36}$$

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

$$\sqrt{\left(q-\frac{19}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Whakarūnātia.

$$q-\frac{19}{6}=\frac{13}{6} q-\frac{19}{6}=-\frac{13}{6}$$

Me tāpiri \frac{19}{6} ki ngā taha e rua o te whārite.

$$q=\frac{16}{3} q=1$$