Whakaoti i te (x^2+3x-2)(x^2+3x+4)=16

$$( x ^ { 2 } + 3 x - 2 ) ( x ^ { 2 } + 3 x + 4 ) = 16$$

Whakaoti mō x (complex solution)

$$x=1$$

$$x=-4$$

$$x=\frac{-3+\sqrt{15}i}{2}\approx -1.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i-3}{2}\approx -1.5-1.936491673i$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}+3x-2 ki te x^{2}+3x+4 ka whakakotahi i ngā kupu rite.

$$x^{4}+6x^{3}+11x^{2}+6x-8=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{4}+6x^{3}+11x^{2}+6x-8-16=0$$

Tangohia te 16 i te -8, ka -24.

$$x^{4}+6x^{3}+11x^{2}+6x-24=0$$

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 -24, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{4}+6x^{3}+11x^{2}+6x-24 ki te x-1, kia riro ko x^{3}+7x^{2}+18x+24. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{3}+7x^{2}+18x+24=0$$

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

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=-4$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}+7x^{2}+18x+24 ki te x+4, kia riro ko x^{2}+3x+6. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 3 mō te b, me te 6 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 6}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{-15}}{2}$$

Whakaotia te whārite x^{2}+3x+6=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{-\sqrt{15}i-3}{2} x=\frac{-3+\sqrt{15}i}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=1 x=-4 x=\frac{-\sqrt{15}i-3}{2} x=\frac{-3+\sqrt{15}i}{2}$$

Whakaoti mō x

$$x=-4$$

$$x=1$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te x^{2}+3x-2 ki te x^{2}+3x+4 ka whakakotahi i ngā kupu rite.

$$x^{4}+6x^{3}+11x^{2}+6x-8=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{4}+6x^{3}+11x^{2}+6x-8-16=0$$

Tangohia te 16 i te -8, ka -24.

$$x^{4}+6x^{3}+11x^{2}+6x-24=0$$

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 -24, ā, ka wehea e q te whakarea arahanga 1. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{4}+6x^{3}+11x^{2}+6x-24 ki te x-1, kia riro ko x^{3}+7x^{2}+18x+24. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$x^{3}+7x^{2}+18x+24=0$$

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

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=-4$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te x^{3}+7x^{2}+18x+24 ki te x+4, kia riro ko x^{2}+3x+6. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 3 mō te b, me te 6 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 6}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{-15}}{2}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā.

$$x\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$x=1 x=-4$$