Whakaoti i te 5p^2+3p/1+p=4

$$\frac { 5 p ^ { 2 } + 3 p } { 1 + p } = 4$$

Whakaoti mō p

$$p=-\frac{4}{5}=-0.8$$

$$p=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe p ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te p+1.

$$5p^{2}+3p=4p+4$$

Tangohia te 4p mai i ngā taha e rua.

$$5p^{2}+3p-4p=4$$

Pahekotia te 3p me -4p, ka -p.

$$5p^{2}-p=4$$

Tangohia te 4 mai i ngā taha e rua.

$$5p^{2}-p-4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 5p^{2}+ap+bp-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=5\left(-4\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-20=-19 2-10=-8 4-5=-1$$

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

$$a=-5 b=4$$

Tuhia anō te 5p^{2}-p-4 hei \left(5p^{2}-5p\right)+\left(4p-4\right).

$$\left(5p^{2}-5p\right)+\left(4p-4\right)$$

Tauwehea te 5p i te tuatahi me te 4 i te rōpū tuarua.

$$5p\left(p-1\right)+4\left(p-1\right)$$

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

$$\left(p-1\right)\left(5p+4\right)$$

Hei kimi otinga whārite, me whakaoti te p-1=0 me te 5p+4=0.

$$p=1 p=-\frac{4}{5}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe p ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te p+1.

$$5p^{2}+3p=4p+4$$

Tangohia te 4p mai i ngā taha e rua.

$$5p^{2}+3p-4p=4$$

Pahekotia te 3p me -4p, ka -p.

$$5p^{2}-p=4$$

Tangohia te 4 mai i ngā taha e rua.

$$5p^{2}-p-4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -1 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-4\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$p=\frac{-\left(-1\right)±\sqrt{1-20\left(-4\right)}}{2\times 5}$$

Whakareatia -20 ki te -4.

$$p=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 5}$$

Tāpiri 1 ki te 80.

$$p=\frac{-\left(-1\right)±\sqrt{81}}{2\times 5}$$

Tuhia te pūtakerua o te 81.

$$p=\frac{-\left(-1\right)±9}{2\times 5}$$

Ko te tauaro o -1 ko 1.

$$p=\frac{1±9}{2\times 5}$$

Whakareatia 2 ki te 5.

$$p=\frac{1±9}{10}$$

Nā, me whakaoti te whārite p=\frac{1±9}{10} ina he tāpiri te ±. Tāpiri 1 ki te 9.

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

Whakawehe 10 ki te 10.

$$p=1$$

Nā, me whakaoti te whārite p=\frac{1±9}{10} ina he tango te ±. Tango 9 mai i 1.

$$p=-\frac{8}{10}$$

Whakahekea te hautanga \frac{-8}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$p=-\frac{4}{5}$$

Kua oti te whārite te whakatau.

$$p=1 p=-\frac{4}{5}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe p ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te p+1.

$$5p^{2}+3p=4\left(p+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te p+1.

$$5p^{2}+3p=4p+4$$

Tangohia te 4p mai i ngā taha e rua.

$$5p^{2}+3p-4p=4$$

Pahekotia te 3p me -4p, ka -p.

$$5p^{2}-p=4$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5p^{2}-p}{5}=\frac{4}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$p^{2}-\frac{1}{5}p=\frac{4}{5}$$

Whakawehea te -\frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{10}. Nā, tāpiria te pūrua o te -\frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$p^{2}-\frac{1}{5}p+\left(-\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(-\frac{1}{10}\right)^{2}$$

Pūruatia -\frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$p^{2}-\frac{1}{5}p+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}$$

Tāpiri \frac{4}{5} ki te \frac{1}{100} 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.

$$p^{2}-\frac{1}{5}p+\frac{1}{100}=\frac{81}{100}$$

Tauwehea p^{2}-\frac{1}{5}p+\frac{1}{100}. 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}.

$$\left(p-\frac{1}{10}\right)^{2}=\frac{81}{100}$$

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

$$\sqrt{\left(p-\frac{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Whakarūnātia.

$$p-\frac{1}{10}=\frac{9}{10} p-\frac{1}{10}=-\frac{9}{10}$$

Me tāpiri \frac{1}{10} ki ngā taha e rua o te whārite.

$$p=1 p=-\frac{4}{5}$$