Whakaoti i te 2x^2-10x=6+11sqrt{x^2-5x}

$$2 x ^ { 2 } - 10 x = 6 + 11 \sqrt { x ^ { 2 } - 5 x }$$

Whakaoti mō x

$$x=-4$$

$$x=9$$

Ngā Upane Otinga

Me tango 6 mai i ngā taha e rua o te whārite.

$$2x^{2}-10x-6=11\sqrt{x^{2}-5x}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(2x^{2}-10x-6\right)^{2}=\left(11\sqrt{x^{2}-5x}\right)^{2}$$

Pūrua 2x^{2}-10x-6.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=\left(11\sqrt{x^{2}-5x}\right)^{2}$$

Whakarohaina te \left(11\sqrt{x^{2}-5x}\right)^{2}.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=11^{2}\left(\sqrt{x^{2}-5x}\right)^{2}$$

Tātaihia te 11 mā te pū o 2, kia riro ko 121.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121\left(\sqrt{x^{2}-5x}\right)^{2}$$

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

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121\left(x^{2}-5x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 121 ki te x^{2}-5x.

$$4x^{4}-40x^{3}+76x^{2}+120x+36=121x^{2}-605x$$

Tangohia te 121x^{2} mai i ngā taha e rua.

$$4x^{4}-40x^{3}+76x^{2}+120x+36-121x^{2}=-605x$$

Pahekotia te 76x^{2} me -121x^{2}, ka -45x^{2}.

$$4x^{4}-40x^{3}-45x^{2}+120x+36=-605x$$

Me tāpiri te 605x ki ngā taha e rua.

$$4x^{4}-40x^{3}-45x^{2}+120x+36+605x=0$$

Pahekotia te 120x me 605x, ka 725x.

$$4x^{4}-40x^{3}-45x^{2}+725x+36=0$$

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 36, ā, ka wehea e q te whakarea arahanga 4. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±9,±18,±36,±\frac{9}{2},±3,±6,±12,±\frac{9}{4},±\frac{3}{2},±1,±2,±4,±\frac{3}{4},±\frac{1}{2},±\frac{1}{4}$$

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 4x^{4}-40x^{3}-45x^{2}+725x+36 ki te x+4, kia riro ko 4x^{3}-56x^{2}+179x+9. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$4x^{3}-56x^{2}+179x+9=0$$

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

$$±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=9$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 4x^{3}-56x^{2}+179x+9 ki te x-9, kia riro ko 4x^{2}-20x-1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$4x^{2}-20x-1=0$$

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 -20 mō te b, me te -1 mō te c i te ture pūrua.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$x=\frac{20±4\sqrt{26}}{8}$$

Whakaotia te whārite 4x^{2}-20x-1=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{5-\sqrt{26}}{2} x=\frac{\sqrt{26}+5}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=-4 x=9 x=\frac{5-\sqrt{26}}{2} x=\frac{\sqrt{26}+5}{2}$$

Whakakapia te -4 mō te x i te whārite 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\left(-4\right)^{2}-10\left(-4\right)=6+11\sqrt{\left(-4\right)^{2}-5\left(-4\right)}$$

Whakarūnātia. Ko te uara x=-4 kua ngata te whārite.

$$72=72$$

Whakakapia te 9 mō te x i te whārite 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times 9^{2}-10\times 9=6+11\sqrt{9^{2}-5\times 9}$$

Whakarūnātia. Ko te uara x=9 kua ngata te whārite.

$$72=72$$

Whakakapia te \frac{5-\sqrt{26}}{2} mō te x i te whārite 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times \left(\frac{5-\sqrt{26}}{2}\right)^{2}-10\times \frac{5-\sqrt{26}}{2}=6+11\sqrt{\left(\frac{5-\sqrt{26}}{2}\right)^{2}-5\times \frac{5-\sqrt{26}}{2}}$$

Whakarūnātia. Ko te uara x=\frac{5-\sqrt{26}}{2} kāore e ngata ana ki te whārite.

$$\frac{1}{2}=\frac{23}{2}$$

Whakakapia te \frac{\sqrt{26}+5}{2} mō te x i te whārite 2x^{2}-10x=6+11\sqrt{x^{2}-5x}.

$$2\times \left(\frac{\sqrt{26}+5}{2}\right)^{2}-10\times \frac{\sqrt{26}+5}{2}=6+11\sqrt{\left(\frac{\sqrt{26}+5}{2}\right)^{2}-5\times \frac{\sqrt{26}+5}{2}}$$

Whakarūnātia. Ko te uara x=\frac{\sqrt{26}+5}{2} kāore e ngata ana ki te whārite.

$$\frac{1}{2}=\frac{23}{2}$$

Rārangihia ngā rongoā katoa o 2x^{2}-10x-6=11\sqrt{x^{2}-5x}.

$$x=-4 x=9$$