Whakaoti i te u^3-8u^2+11u+20=0

$$u ^ { 3 } - 8 u ^ { 2 } + 11 u + 20 = 0$$

Whakaoti mō u

$$u=4$$

$$u=5$$

$$u=-1$$

Ngā Upane Otinga

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 20, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$u=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te u-k o te pūrau mō ia pūtake k. Whakawehea te u^{3}-8u^{2}+11u+20 ki te u+1, kia riro ko u^{2}-9u+20. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$u^{2}-9u+20=0$$

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 -9 mō te b, me te 20 mō te c i te ture pūrua.

$$u=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 1\times 20}}{2}$$

Mahia ngā tātaitai.

$$u=\frac{9±1}{2}$$

Whakaotia te whārite u^{2}-9u+20=0 ina he tōrunga te ±, ina he tōraro te ±.

$$u=4 u=5$$

Rārangitia ngā otinga katoa i kitea.

$$u=-1 u=4 u=5$$