Whakaoti i te (x^2-4x+3)*sqrt{2x^2-5x+2}=0

$$( x ^ { 2 } - 4 x + 3 ) \cdot \sqrt { 2 x ^ { 2 } - 5 x + 2 } = 0$$

Whakaoti mō x (complex solution)

$$x=1$$

$$x=3$$

$$x=2$$

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

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-4x+3 ki te \sqrt{2x^{2}-5x+2}.

$$x^{2}\sqrt{2x^{2}-5x+2}-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}=0$$

Me tango -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2} mai i ngā taha e rua o te whārite.

$$x^{2}\sqrt{2x^{2}-5x+2}=-\left(-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}\right)$$

Hei kimi i te tauaro o -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}, kimihia te tauaro o ia taurangi.

$$x^{2}\sqrt{2x^{2}-5x+2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakarohaina te \left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}.

$$\left(x^{2}\right)^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 2 me te 2 kia riro ai te 4.

$$x^{4}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$x^{4}\left(2x^{2}-5x+2\right)=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{4} ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\sqrt{2x^{2}-5x+2}\sqrt{2x^{2}-5x+2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakareatia te \sqrt{2x^{2}-5x+2} ki te \sqrt{2x^{2}-5x+2}, ka \left(\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(2x^{2}-5x+2\right)-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 16x^{2} ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(2x^{2}-5x+2\right)+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te -24x ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-48x^{3}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Pahekotia te -80x^{3} me -48x^{3}, ka -128x^{3}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+32x^{2}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Pahekotia te 32x^{2} me 120x^{2}, ka 152x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(2x^{2}-5x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9 ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+18x^{2}-45x+18$$

Pahekotia te 152x^{2} me 18x^{2}, ka 170x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-48x-45x+18$$

Pahekotia te -48x me -45x, ka -93x.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-93x+18$$

Tangohia te 32x^{4} mai i ngā taha e rua.

$$2x^{6}-5x^{5}+2x^{4}-32x^{4}=-128x^{3}+170x^{2}-93x+18$$

Pahekotia te 2x^{4} me -32x^{4}, ka -30x^{4}.

$$2x^{6}-5x^{5}-30x^{4}=-128x^{3}+170x^{2}-93x+18$$

Me tāpiri te 128x^{3} ki ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}=170x^{2}-93x+18$$

Tangohia te 170x^{2} mai i ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}=-93x+18$$

Me tāpiri te 93x ki ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x=18$$

Tangohia te 18 mai i ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18=0$$

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 -18, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18 ki te x-1, kia riro ko 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18=0$$

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 18, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18 ki te x-2, kia riro ko 2x^{4}+x^{3}-31x^{2}+33x-9. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{4}+x^{3}-31x^{2}+33x-9=0$$

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 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{4}+x^{3}-31x^{2}+33x-9 ki te x-3, kia riro ko 2x^{3}+7x^{2}-10x+3. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{3}+7x^{2}-10x+3=0$$

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 3, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

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

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{3}+7x^{2}-10x+3 ki te 2\left(x-\frac{1}{2}\right)=2x-1, kia riro ko x^{2}+4x-3. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 4 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Whakaotia te whārite x^{2}+4x-3=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Rārangitia ngā otinga katoa i kitea.

$$x=1 x=2 x=3 x=\frac{1}{2} x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Whakakapia te 1 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(1^{2}-4+3\right)\sqrt{2\times 1^{2}-5+2}=0$$

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$0=0$$

Whakakapia te 2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(2^{2}-4\times 2+3\right)\sqrt{2\times 2^{2}-5\times 2+2}=0$$

Whakarūnātia. Ko te uara x=2 kua ngata te whārite.

$$0=0$$

Whakakapia te 3 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(3^{2}-4\times 3+3\right)\sqrt{2\times 3^{2}-5\times 3+2}=0$$

Whakarūnātia. Ko te uara x=3 kua ngata te whārite.

$$0=0$$

Whakakapia te \frac{1}{2} mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{2}+3\right)\sqrt{2\times \left(\frac{1}{2}\right)^{2}-5\times \frac{1}{2}+2}=0$$

Whakarūnātia. Ko te uara x=\frac{1}{2} kua ngata te whārite.

$$0=0$$

Whakakapia te -\sqrt{7}-2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(-\sqrt{7}-2\right)^{2}-4\left(-\sqrt{7}-2\right)+3\right)\sqrt{2\left(-\sqrt{7}-2\right)^{2}-5\left(-\sqrt{7}-2\right)+2}=0$$

Whakarūnātia. Ko te uara x=-\sqrt{7}-2 kāore e ngata ana ki te whārite.

$$\left(63720+24084\times 7^{\frac{1}{2}}\right)^{\frac{1}{2}}=0$$

Whakakapia te \sqrt{7}-2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\sqrt{7}-2\right)^{2}-4\left(\sqrt{7}-2\right)+3\right)\sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2}=0$$

Whakarūnātia. Ko te uara x=\sqrt{7}-2 kāore e ngata ana ki te whārite.

$$i\left(-\left(63720-24084\times 7^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}=0$$

Rārangihia ngā rongoā katoa o \sqrt{2x^{2}-5x+2}x^{2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}.

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

Whakaoti mō x

$$x=3$$

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

$$x=2$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-4x+3 ki te \sqrt{2x^{2}-5x+2}.

$$x^{2}\sqrt{2x^{2}-5x+2}-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}=0$$

Me tango -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2} mai i ngā taha e rua o te whārite.

$$x^{2}\sqrt{2x^{2}-5x+2}=-\left(-4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}\right)$$

Hei kimi i te tauaro o -4x\sqrt{2x^{2}-5x+2}+3\sqrt{2x^{2}-5x+2}, kimihia te tauaro o ia taurangi.

$$x^{2}\sqrt{2x^{2}-5x+2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakarohaina te \left(x^{2}\sqrt{2x^{2}-5x+2}\right)^{2}.

$$\left(x^{2}\right)^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 2 me te 2 kia riro ai te 4.

$$x^{4}\left(\sqrt{2x^{2}-5x+2}\right)^{2}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$x^{4}\left(2x^{2}-5x+2\right)=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{4} ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=\left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\sqrt{2x^{2}-5x+2}\sqrt{2x^{2}-5x+2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakareatia te \sqrt{2x^{2}-5x+2} ki te \sqrt{2x^{2}-5x+2}, ka \left(\sqrt{2x^{2}-5x+2}\right)^{2}.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(\sqrt{2x^{2}-5x+2}\right)^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=16x^{2}\left(2x^{2}-5x+2\right)-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te 16x^{2} ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(\sqrt{2x^{2}-5x+2}\right)^{2}+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-24x\left(2x^{2}-5x+2\right)+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te -24x ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-80x^{3}+32x^{2}-48x^{3}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Pahekotia te -80x^{3} me -48x^{3}, ka -128x^{3}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+32x^{2}+120x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Pahekotia te 32x^{2} me 120x^{2}, ka 152x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(\sqrt{2x^{2}-5x+2}\right)^{2}$$

Tātaihia te \sqrt{2x^{2}-5x+2} mā te pū o 2, kia riro ko 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+9\left(2x^{2}-5x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9 ki te 2x^{2}-5x+2.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+152x^{2}-48x+18x^{2}-45x+18$$

Pahekotia te 152x^{2} me 18x^{2}, ka 170x^{2}.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-48x-45x+18$$

Pahekotia te -48x me -45x, ka -93x.

$$2x^{6}-5x^{5}+2x^{4}=32x^{4}-128x^{3}+170x^{2}-93x+18$$

Tangohia te 32x^{4} mai i ngā taha e rua.

$$2x^{6}-5x^{5}+2x^{4}-32x^{4}=-128x^{3}+170x^{2}-93x+18$$

Pahekotia te 2x^{4} me -32x^{4}, ka -30x^{4}.

$$2x^{6}-5x^{5}-30x^{4}=-128x^{3}+170x^{2}-93x+18$$

Me tāpiri te 128x^{3} ki ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}=170x^{2}-93x+18$$

Tangohia te 170x^{2} mai i ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}=-93x+18$$

Me tāpiri te 93x ki ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x=18$$

Tangohia te 18 mai i ngā taha e rua.

$$2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18=0$$

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 -18, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{6}-5x^{5}-30x^{4}+128x^{3}-170x^{2}+93x-18 ki te x-1, kia riro ko 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18=0$$

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 18, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±18,±\frac{9}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 2x^{5}-3x^{4}-33x^{3}+95x^{2}-75x+18 ki te x-2, kia riro ko 2x^{4}+x^{3}-31x^{2}+33x-9. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{4}+x^{3}-31x^{2}+33x-9=0$$

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 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{4}+x^{3}-31x^{2}+33x-9 ki te x-3, kia riro ko 2x^{3}+7x^{2}-10x+3. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{3}+7x^{2}-10x+3=0$$

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 3, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

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

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{3}+7x^{2}-10x+3 ki te 2\left(x-\frac{1}{2}\right)=2x-1, kia riro ko x^{2}+4x-3. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 4 mō te b, me te -3 mō te c i te ture pūrua.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-3\right)}}{2}$$

Mahia ngā tātaitai.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Whakaotia te whārite x^{2}+4x-3=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Rārangitia ngā otinga katoa i kitea.

$$x=1 x=2 x=3 x=\frac{1}{2} x=-\sqrt{7}-2 x=\sqrt{7}-2$$

Whakakapia te 1 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0. Te kīanga \sqrt{2\times 1^{2}-5+2} kia kore e tautuhitia nā te mea kāore te radicand e noho tōraro.

$$\left(1^{2}-4+3\right)\sqrt{2\times 1^{2}-5+2}=0$$

Whakakapia te 2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(2^{2}-4\times 2+3\right)\sqrt{2\times 2^{2}-5\times 2+2}=0$$

Whakarūnātia. Ko te uara x=2 kua ngata te whārite.

$$0=0$$

Whakakapia te 3 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(3^{2}-4\times 3+3\right)\sqrt{2\times 3^{2}-5\times 3+2}=0$$

Whakarūnātia. Ko te uara x=3 kua ngata te whārite.

$$0=0$$

Whakakapia te \frac{1}{2} mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{2}+3\right)\sqrt{2\times \left(\frac{1}{2}\right)^{2}-5\times \frac{1}{2}+2}=0$$

Whakarūnātia. Ko te uara x=\frac{1}{2} kua ngata te whārite.

$$0=0$$

Whakakapia te -\sqrt{7}-2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0.

$$\left(\left(-\sqrt{7}-2\right)^{2}-4\left(-\sqrt{7}-2\right)+3\right)\sqrt{2\left(-\sqrt{7}-2\right)^{2}-5\left(-\sqrt{7}-2\right)+2}=0$$

Whakarūnātia. Ko te uara x=-\sqrt{7}-2 kāore e ngata ana ki te whārite.

$$\left(63720+24084\times 7^{\frac{1}{2}}\right)^{\frac{1}{2}}=0$$

Whakakapia te \sqrt{7}-2 mō te x i te whārite \left(x^{2}-4x+3\right)\sqrt{2x^{2}-5x+2}=0. Te kīanga \sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2} kia kore e tautuhitia nā te mea kāore te radicand e noho tōraro.

$$\left(\left(\sqrt{7}-2\right)^{2}-4\left(\sqrt{7}-2\right)+3\right)\sqrt{2\left(\sqrt{7}-2\right)^{2}-5\left(\sqrt{7}-2\right)+2}=0$$

Rārangihia ngā rongoā katoa o \sqrt{2x^{2}-5x+2}x^{2}=4x\sqrt{2x^{2}-5x+2}-3\sqrt{2x^{2}-5x+2}.

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