Whakaoti i te 3/p-p-1/p=3p

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

Whakaoti mō p

$$p = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$p=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe p ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p.

$$3-\left(p-1\right)=3pp$$

Whakareatia te p ki te p, ka p^{2}.

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

Hei kimi i te tauaro o p-1, kimihia te tauaro o ia taurangi.

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

Ko te tauaro o -1 ko 1.

$$3-p+1=3p^{2}$$

Tāpirihia te 3 ki te 1, ka 4.

$$4-p=3p^{2}$$

Tangohia te 3p^{2} mai i ngā taha e rua.

$$4-p-3p^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-3p^{2}-p+4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -3p^{2}+ap+bp+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-3\times 4=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

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

$$a=3 b=-4$$

Tuhia anō te -3p^{2}-p+4 hei \left(-3p^{2}+3p\right)+\left(-4p+4\right).

$$\left(-3p^{2}+3p\right)+\left(-4p+4\right)$$

Tauwehea te 3p i te tuatahi me te 4 i te rōpū tuarua.

$$3p\left(-p+1\right)+4\left(-p+1\right)$$

Whakatauwehea atu te kīanga pātahi -p+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-p+1\right)\left(3p+4\right)$$

Hei kimi otinga whārite, me whakaoti te -p+1=0 me te 3p+4=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe p ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p.

$$3-\left(p-1\right)=3pp$$

Whakareatia te p ki te p, ka p^{2}.

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

Hei kimi i te tauaro o p-1, kimihia te tauaro o ia taurangi.

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

Ko te tauaro o -1 ko 1.

$$3-p+1=3p^{2}$$

Tāpirihia te 3 ki te 1, ka 4.

$$4-p=3p^{2}$$

Tangohia te 3p^{2} mai i ngā taha e rua.

$$4-p-3p^{2}=0$$

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.

$$-3p^{2}-p+4=0$$

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

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

Whakareatia -4 ki te -3.

$$p=\frac{-\left(-1\right)±\sqrt{1+12\times 4}}{2\left(-3\right)}$$

Whakareatia 12 ki te 4.

$$p=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-3\right)}$$

Tāpiri 1 ki te 48.

$$p=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 49.

$$p=\frac{-\left(-1\right)±7}{2\left(-3\right)}$$

Ko te tauaro o -1 ko 1.

$$p=\frac{1±7}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$p=\frac{1±7}{-6}$$

Nā, me whakaoti te whārite p=\frac{1±7}{-6} ina he tāpiri te ±. Tāpiri 1 ki te 7.

$$p=\frac{8}{-6}$$

Whakahekea te hautanga \frac{8}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite p=\frac{1±7}{-6} ina he tango te ±. Tango 7 mai i 1.

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

Whakawehe -6 ki te -6.

$$p=1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe p ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p.

$$3-\left(p-1\right)=3pp$$

Whakareatia te p ki te p, ka p^{2}.

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

Hei kimi i te tauaro o p-1, kimihia te tauaro o ia taurangi.

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

Ko te tauaro o -1 ko 1.

$$3-p+1=3p^{2}$$

Tāpirihia te 3 ki te 1, ka 4.

$$4-p=3p^{2}$$

Tangohia te 3p^{2} mai i ngā taha e rua.

$$4-p-3p^{2}=0$$

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-p-3p^{2}=-4$$

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.

$$-3p^{2}-p=-4$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3p^{2}-p}{-3}=-\frac{4}{-3}$$

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

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

Whakawehe -1 ki te -3.

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

Whakawehe -4 ki te -3.

$$p^{2}+\frac{1}{3}p=\frac{4}{3}$$

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.

$$p^{2}+\frac{1}{3}p+\left(\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(\frac{1}{6}\right)^{2}$$

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

$$p^{2}+\frac{1}{3}p+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Tāpiri \frac{4}{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.

$$p^{2}+\frac{1}{3}p+\frac{1}{36}=\frac{49}{36}$$

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

$$\left(p+\frac{1}{6}\right)^{2}=\frac{49}{36}$$

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

$$\sqrt{\left(p+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Whakarūnātia.

$$p+\frac{1}{6}=\frac{7}{6} p+\frac{1}{6}=-\frac{7}{6}$$

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

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