Whakaoti i te p^2-9p+18=0

$$p ^ { 2 } - 9 p + 18 = 0$$

Whakaoti mō p

$$p=3$$

$$p=6$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te p^{2}-9p+18 mā te whakamahi i te tātai p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

Ko te otinga te takirua ka hoatu i te tapeke -9.

$$a=-6 b=-3$$

Me tuhi anō te kīanga whakatauwehe \left(p+a\right)\left(p+b\right) mā ngā uara i tātaihia.

$$\left(p-6\right)\left(p-3\right)$$

Hei kimi otinga whārite, me whakaoti te p-6=0 me te p-3=0.

$$p=6 p=3$$

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 p^{2}+ap+bp+18. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=1\times 18=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

Ko te otinga te takirua ka hoatu i te tapeke -9.

$$a=-6 b=-3$$

Tuhia anō te p^{2}-9p+18 hei \left(p^{2}-6p\right)+\left(-3p+18\right).

$$\left(p^{2}-6p\right)+\left(-3p+18\right)$$

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

$$p\left(p-6\right)-3\left(p-6\right)$$

Whakatauwehea atu te kīanga pātahi p-6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(p-6\right)\left(p-3\right)$$

Hei kimi otinga whārite, me whakaoti te p-6=0 me te p-3=0.

$$p=6 p=3$$

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.

$$p^{2}-9p+18=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -9 mō b, me 18 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18}}{2}$$

Pūrua -9.

$$p=\frac{-\left(-9\right)±\sqrt{81-4\times 18}}{2}$$

Whakareatia -4 ki te 18.

$$p=\frac{-\left(-9\right)±\sqrt{81-72}}{2}$$

Tāpiri 81 ki te -72.

$$p=\frac{-\left(-9\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$p=\frac{-\left(-9\right)±3}{2}$$

Ko te tauaro o -9 ko 9.

$$p=\frac{9±3}{2}$$

Nā, me whakaoti te whārite p=\frac{9±3}{2} ina he tāpiri te ±. Tāpiri 9 ki te 3.

$$p=\frac{12}{2}$$

Whakawehe 12 ki te 2.

$$p=6$$

Nā, me whakaoti te whārite p=\frac{9±3}{2} ina he tango te ±. Tango 3 mai i 9.

$$p=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$p=3$$

Kua oti te whārite te whakatau.

$$p=6 p=3$$

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.

$$p^{2}-9p+18=0$$

Me tango 18 mai i ngā taha e rua o te whārite.

$$p^{2}-9p+18-18=-18$$

Mā te tango i te 18 i a ia ake anō ka toe ko te 0.

$$p^{2}-9p=-18$$

Whakawehea te -9, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{9}{2}. Nā, tāpiria te pūrua o te -\frac{9}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$p^{2}-9p+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Pūruatia -\frac{9}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$p^{2}-9p+\frac{81}{4}=-18+\frac{81}{4}$$

Tāpiri -18 ki te \frac{81}{4}.

$$p^{2}-9p+\frac{81}{4}=\frac{9}{4}$$

Tauwehea p^{2}-9p+\frac{81}{4}. 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}.

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

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

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

Whakarūnātia.

$$p-\frac{9}{2}=\frac{3}{2} p-\frac{9}{2}=-\frac{3}{2}$$

Me tāpiri \frac{9}{2} ki ngā taha e rua o te whārite.

$$p=6 p=3$$