Whakaoti i te 12x^3+8x^2=x+1

$$12 x ^ { 3 } + 8 x ^ { 2 } = x + 1$$

Whakaoti mō x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{3}\approx 0.333333333$$

Ngā Upane Otinga

Tangohia te x mai i ngā taha e rua.

$$12x^{3}+8x^{2}-x=1$$

Tangohia te 1 mai i ngā taha e rua.

$$12x^{3}+8x^{2}-x-1=0$$

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

$$±\frac{1}{12},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

$$x=\frac{1}{3}$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 12x^{3}+8x^{2}-x-1 ki te 3\left(x-\frac{1}{3}\right)=3x-1, kia riro ko 4x^{2}+4x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$4x^{2}+4x+1=0$$

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 4 mō te a, te 4 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\times 1}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{-4±0}{8}$$

He ōrite ngā whakatau.

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

Rārangitia ngā otinga katoa i kitea.

$$x=\frac{1}{3} x=-\frac{1}{2}$$