Whakaoti i te 10x^2+x^3-6=3

$$10 x ^ { 2 } + x ^ { 3 } - 6 = 3$$

Whakaoti mō x

$$x=\frac{-3\sqrt{13}-9}{2}\approx -9.908326913$$

$$x=-1$$

$$x=\frac{3\sqrt{13}-9}{2}\approx 0.908326913$$

Ngā Upane Otinga

Tangohia te 3 mai i ngā taha e rua.

$$10x^{2}+x^{3}-6-3=0$$

Tangohia te 3 i te -6, ka -9.

$$10x^{2}+x^{3}-9=0$$

Hurinahatia te whārite ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$x^{3}+10x^{2}-9=0$$

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 -9, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}+10x^{2}-9 ki te x+1, kia riro ko x^{2}+9x-9. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{2}+9x-9=0$$

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 -9 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{-9±3\sqrt{13}}{2}$$

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

$$x=\frac{-3\sqrt{13}-9}{2} x=\frac{3\sqrt{13}-9}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=-1 x=\frac{-3\sqrt{13}-9}{2} x=\frac{3\sqrt{13}-9}{2}$$