Whakaoti i te 3n^2+3n+1=1141

$$3 n ^ { 2 } + 3 n + 1 = 1141$$

Whakaoti mō n

$$n=-20$$

$$n=19$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 1141 mai i ngā taha e rua.

$$3n^{2}+3n+1-1141=0$$

Tangohia te 1141 i te 1, ka -1140.

$$3n^{2}+3n-1140=0$$

Whakawehea ngā taha e rua ki te 3.

$$n^{2}+n-380=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei n^{2}+an+bn-380. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=1\left(-380\right)=-380$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -380.

$$-1,380 -2,190 -4,95 -5,76 -10,38 -19,20$$

Tātaihia te tapeke mō ia takirua.

$$-1+380=379 -2+190=188 -4+95=91 -5+76=71 -10+38=28 -19+20=1$$

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

$$a=-19 b=20$$

Tuhia anō te n^{2}+n-380 hei \left(n^{2}-19n\right)+\left(20n-380\right).

$$\left(n^{2}-19n\right)+\left(20n-380\right)$$

Tauwehea te n i te tuatahi me te 20 i te rōpū tuarua.

$$n\left(n-19\right)+20\left(n-19\right)$$

Whakatauwehea atu te kīanga pātahi n-19 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(n-19\right)\left(n+20\right)$$

Hei kimi otinga whārite, me whakaoti te n-19=0 me te n+20=0.

$$n=19 n=-20$$

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.

$$3n^{2}+3n+1=1141$$

Me tango 1141 mai i ngā taha e rua o te whārite.

$$3n^{2}+3n+1-1141=1141-1141$$

Mā te tango i te 1141 i a ia ake anō ka toe ko te 0.

$$3n^{2}+3n+1-1141=0$$

Tango 1141 mai i 1.

$$3n^{2}+3n-1140=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 3 mō b, me -1140 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\times 3\left(-1140\right)}}{2\times 3}$$

Pūrua 3.

$$n=\frac{-3±\sqrt{9-4\times 3\left(-1140\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$n=\frac{-3±\sqrt{9-12\left(-1140\right)}}{2\times 3}$$

Whakareatia -12 ki te -1140.

$$n=\frac{-3±\sqrt{9+13680}}{2\times 3}$$

Tāpiri 9 ki te 13680.

$$n=\frac{-3±\sqrt{13689}}{2\times 3}$$

Tuhia te pūtakerua o te 13689.

$$n=\frac{-3±117}{2\times 3}$$

Whakareatia 2 ki te 3.

$$n=\frac{-3±117}{6}$$

Nā, me whakaoti te whārite n=\frac{-3±117}{6} ina he tāpiri te ±. Tāpiri -3 ki te 117.

$$n=\frac{114}{6}$$

Whakawehe 114 ki te 6.

$$n=19$$

Nā, me whakaoti te whārite n=\frac{-3±117}{6} ina he tango te ±. Tango 117 mai i -3.

$$n=-\frac{120}{6}$$

Whakawehe -120 ki te 6.

$$n=-20$$

Kua oti te whārite te whakatau.

$$n=19 n=-20$$

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.

$$3n^{2}+3n+1=1141$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$3n^{2}+3n+1-1=1141-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$3n^{2}+3n=1141-1$$

Tango 1 mai i 1141.

$$3n^{2}+3n=1140$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3n^{2}+3n}{3}=\frac{1140}{3}$$

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

$$n^{2}+\frac{3}{3}n=\frac{1140}{3}$$

Whakawehe 3 ki te 3.

$$n^{2}+n=\frac{1140}{3}$$

Whakawehe 1140 ki te 3.

$$n^{2}+n=380$$

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

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=380+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$n^{2}+n+\frac{1}{4}=380+\frac{1}{4}$$

Tāpiri 380 ki te \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{1521}{4}$$

Tauwehea n^{2}+n+\frac{1}{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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{1521}{4}$$

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

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1521}{4}}$$

Whakarūnātia.

$$n+\frac{1}{2}=\frac{39}{2} n+\frac{1}{2}=-\frac{39}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$n=19 n=-20$$