Whakaoti i te 3p^2-8p+5=0

$$3 p ^ { 2 } - 8 p + 5 = 0$$

Whakaoti mō p

$$p=1$$

$$p = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

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 3p^{2}+ap+bp+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-15=-16 -3-5=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-5 b=-3$$

Tuhia anō te 3p^{2}-8p+5 hei \left(3p^{2}-5p\right)+\left(-3p+5\right).

$$\left(3p^{2}-5p\right)+\left(-3p+5\right)$$

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

$$p\left(3p-5\right)-\left(3p-5\right)$$

Whakatauwehea atu te kīanga pātahi 3p-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3p-5\right)\left(p-1\right)$$

Hei kimi otinga whārite, me whakaoti te 3p-5=0 me te p-1=0.

$$p=\frac{5}{3} p=1$$

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.

$$3p^{2}-8p+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -8 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Pūrua -8.

$$p=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 5}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$p=\frac{-\left(-8\right)±\sqrt{64-12\times 5}}{2\times 3}$$

Whakareatia -12 ki te 5.

$$p=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 3}$$

Tāpiri 64 ki te -60.

$$p=\frac{-\left(-8\right)±\sqrt{4}}{2\times 3}$$

Tuhia te pūtakerua o te 4.

$$p=\frac{-\left(-8\right)±2}{2\times 3}$$

Ko te tauaro o -8 ko 8.

$$p=\frac{8±2}{2\times 3}$$

Whakareatia 2 ki te 3.

$$p=\frac{8±2}{6}$$

Nā, me whakaoti te whārite p=\frac{8±2}{6} ina he tāpiri te ±. Tāpiri 8 ki te 2.

$$p=\frac{10}{6}$$

Whakahekea te hautanga \frac{10}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$p=\frac{5}{3}$$

Nā, me whakaoti te whārite p=\frac{8±2}{6} ina he tango te ±. Tango 2 mai i 8.

$$p=\frac{6}{6}$$

Whakawehe 6 ki te 6.

$$p=1$$

Kua oti te whārite te whakatau.

$$p=\frac{5}{3} p=1$$

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.

$$3p^{2}-8p+5=0$$

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

$$3p^{2}-8p+5-5=-5$$

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

$$3p^{2}-8p=-5$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3p^{2}-8p}{3}=-\frac{5}{3}$$

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

$$p^{2}-\frac{8}{3}p=-\frac{5}{3}$$

Whakawehea te -\frac{8}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{4}{3}. Nā, tāpiria te pūrua o te -\frac{4}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$p^{2}-\frac{8}{3}p+\left(-\frac{4}{3}\right)^{2}=-\frac{5}{3}+\left(-\frac{4}{3}\right)^{2}$$

Pūruatia -\frac{4}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$p^{2}-\frac{8}{3}p+\frac{16}{9}=-\frac{5}{3}+\frac{16}{9}$$

Tāpiri -\frac{5}{3} ki te \frac{16}{9} 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.

$$p^{2}-\frac{8}{3}p+\frac{16}{9}=\frac{1}{9}$$

Tauwehea p^{2}-\frac{8}{3}p+\frac{16}{9}. 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}.

$$\left(p-\frac{4}{3}\right)^{2}=\frac{1}{9}$$

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

$$\sqrt{\left(p-\frac{4}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Whakarūnātia.

$$p-\frac{4}{3}=\frac{1}{3} p-\frac{4}{3}=-\frac{1}{3}$$

Me tāpiri \frac{4}{3} ki ngā taha e rua o te whārite.

$$p=\frac{5}{3} p=1$$