Whakaoti i te x^3-6x^2-2x+12=0

$${ x }^{ 3 } -6 { x }^{ 2 } -2x+12=0$$

Whakaoti mō x

$$x=-\sqrt{2}\approx -1.414213562$$

$$x=6$$

$$x=\sqrt{2}\approx 1.414213562$$

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

$$±12,±6,±4,±3,±2,±1$$

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.

$$x=6$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}-6x^{2}-2x+12 ki te x-6, kia riro ko x^{2}-2. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}-2=0$$

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 0 mō te b, me te -2 mō te c i te ture pūrua.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-2\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{0±2\sqrt{2}}{2}$$

Whakaotia te whārite x^{2}-2=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\sqrt{2} x=\sqrt{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=6 x=-\sqrt{2} x=\sqrt{2}$$