Whakaoti i te x^3+3x^2-3x-10=0

$${ x }^{ 3 } +3 { x }^{ 2 } -3x-10=0$$

Whakaoti mō x

$$x = \frac{\sqrt{21} - 1}{2} \approx 1.791287847$$

$$x=\frac{-\sqrt{21}-1}{2}\approx -2.791287847$$

$$x=-2$$

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 -10, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±10,±5,±2,±1$$

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.

$$x=-2$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}+3x^{2}-3x-10 ki te x+2, kia riro ko x^{2}+x-5. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}+x-5=0$$

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 1 mō te b, me te -5 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-5\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{21}}{2}$$

Whakaotia te whārite x^{2}+x-5=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{-\sqrt{21}-1}{2} x=\frac{\sqrt{21}-1}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=-2 x=\frac{-\sqrt{21}-1}{2} x=\frac{\sqrt{21}-1}{2}$$