Whakaoti i te n^2-33n+260=0

$$n ^ { 2 } - 33 n + 260 = 0$$

Whakaoti mō n

$$n=13$$

$$n=20$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te n^{2}-33n+260 mā te whakamahi i te tātai n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-33 ab=260$$

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 260.

$$-1,-260 -2,-130 -4,-65 -5,-52 -10,-26 -13,-20$$

Tātaihia te tapeke mō ia takirua.

$$-1-260=-261 -2-130=-132 -4-65=-69 -5-52=-57 -10-26=-36 -13-20=-33$$

Ko te otinga te takirua ka hoatu i te tapeke -33.

$$a=-20 b=-13$$

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

$$\left(n-20\right)\left(n-13\right)$$

Hei kimi otinga whārite, me whakaoti te n-20=0 me te n-13=0.

$$n=20 n=13$$

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 n^{2}+an+bn+260. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-33 ab=1\times 260=260$$

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 260.

$$-1,-260 -2,-130 -4,-65 -5,-52 -10,-26 -13,-20$$

Tātaihia te tapeke mō ia takirua.

$$-1-260=-261 -2-130=-132 -4-65=-69 -5-52=-57 -10-26=-36 -13-20=-33$$

Ko te otinga te takirua ka hoatu i te tapeke -33.

$$a=-20 b=-13$$

Tuhia anō te n^{2}-33n+260 hei \left(n^{2}-20n\right)+\left(-13n+260\right).

$$\left(n^{2}-20n\right)+\left(-13n+260\right)$$

Tauwehea te n i te tuatahi me te -13 i te rōpū tuarua.

$$n\left(n-20\right)-13\left(n-20\right)$$

Whakatauwehea atu te kīanga pātahi n-20 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(n-20\right)\left(n-13\right)$$

Hei kimi otinga whārite, me whakaoti te n-20=0 me te n-13=0.

$$n=20 n=13$$

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.

$$n^{2}-33n+260=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -33 mō b, me 260 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 260}}{2}$$

Pūrua -33.

$$n=\frac{-\left(-33\right)±\sqrt{1089-4\times 260}}{2}$$

Whakareatia -4 ki te 260.

$$n=\frac{-\left(-33\right)±\sqrt{1089-1040}}{2}$$

Tāpiri 1089 ki te -1040.

$$n=\frac{-\left(-33\right)±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$n=\frac{-\left(-33\right)±7}{2}$$

Ko te tauaro o -33 ko 33.

$$n=\frac{33±7}{2}$$

Nā, me whakaoti te whārite n=\frac{33±7}{2} ina he tāpiri te ±. Tāpiri 33 ki te 7.

$$n=\frac{40}{2}$$

Whakawehe 40 ki te 2.

$$n=20$$

Nā, me whakaoti te whārite n=\frac{33±7}{2} ina he tango te ±. Tango 7 mai i 33.

$$n=\frac{26}{2}$$

Whakawehe 26 ki te 2.

$$n=13$$

Kua oti te whārite te whakatau.

$$n=20 n=13$$

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.

$$n^{2}-33n+260=0$$

Me tango 260 mai i ngā taha e rua o te whārite.

$$n^{2}-33n+260-260=-260$$

Mā te tango i te 260 i a ia ake anō ka toe ko te 0.

$$n^{2}-33n=-260$$

Whakawehea te -33, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{33}{2}. Nā, tāpiria te pūrua o te -\frac{33}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$n^{2}-33n+\left(-\frac{33}{2}\right)^{2}=-260+\left(-\frac{33}{2}\right)^{2}$$

Pūruatia -\frac{33}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$n^{2}-33n+\frac{1089}{4}=-260+\frac{1089}{4}$$

Tāpiri -260 ki te \frac{1089}{4}.

$$n^{2}-33n+\frac{1089}{4}=\frac{49}{4}$$

Tauwehea n^{2}-33n+\frac{1089}{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}.

$$\left(n-\frac{33}{2}\right)^{2}=\frac{49}{4}$$

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

$$\sqrt{\left(n-\frac{33}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Whakarūnātia.

$$n-\frac{33}{2}=\frac{7}{2} n-\frac{33}{2}=-\frac{7}{2}$$

Me tāpiri \frac{33}{2} ki ngā taha e rua o te whārite.

$$n=20 n=13$$