Whakaoti i te n^3+n^2+n=1110

$${ n }^{ 3 } + { n }^{ 2 } +n = 1110$$

Whakaoti mō n (complex solution)

$$n=\frac{-\sqrt{323}i-11}{2}\approx -5.5-8.986100378i$$

$$n=10$$

$$n=\frac{-11+\sqrt{323}i}{2}\approx -5.5+8.986100378i$$

Ngā Upane Otinga

Tangohia te 1110 mai i ngā taha e rua.

$$n^{3}+n^{2}+n-1110=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -1110, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±1110,±555,±370,±222,±185,±111,±74,±37,±30,±15,±10,±6,±5,±3,±2,±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$n=10$$

Mā te whakatakotoranga Tauwehe, he tauwehe te n-k o te pūrau mō ia pūtake k. Whakawehea te n^{3}+n^{2}+n-1110 ki te n-10, kia riro ko n^{2}+11n+111. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$n^{2}+11n+111=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te 11 mō te b, me te 111 mō te c i te ture pūrua.

$$n=\frac{-11±\sqrt{11^{2}-4\times 1\times 111}}{2}$$

Mahia ngā tātaitai.

$$n=\frac{-11±\sqrt{-323}}{2}$$

Whakaotia te whārite n^{2}+11n+111=0 ina he tōrunga te ±, ina he tōraro te ±.

$$n=\frac{-\sqrt{323}i-11}{2} n=\frac{-11+\sqrt{323}i}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$n=10 n=\frac{-\sqrt{323}i-11}{2} n=\frac{-11+\sqrt{323}i}{2}$$

Whakaoti mō n

$$n=10$$

Ngā Upane Otinga

Tangohia te 1110 mai i ngā taha e rua.

$$n^{3}+n^{2}+n-1110=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau -1110, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±1110,±555,±370,±222,±185,±111,±74,±37,±30,±15,±10,±6,±5,±3,±2,±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$n=10$$

Mā te whakatakotoranga Tauwehe, he tauwehe te n-k o te pūrau mō ia pūtake k. Whakawehea te n^{3}+n^{2}+n-1110 ki te n-10, kia riro ko n^{2}+11n+111. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$n^{2}+11n+111=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te 11 mō te b, me te 111 mō te c i te ture pūrua.

$$n=\frac{-11±\sqrt{11^{2}-4\times 1\times 111}}{2}$$

Mahia ngā tātaitai.

$$n=\frac{-11±\sqrt{-323}}{2}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā.

$$n\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$n=10$$