Whakaoti i te 4(x^2+1)(2x^2+1)=5(x^2-1)^2

$$4 ( x ^ { 2 } + 1 ) ( 2 x ^ { 2 } + 1 ) = 5 ( x ^ { 2 } - 1 ) ^ { 2 }$$

Whakaoti mō x (complex solution)

$$x=\frac{i\sqrt{6\sqrt{31}+33}}{3}\approx 2.716341211i$$

$$x=-\frac{i\sqrt{6\sqrt{31}+33}}{3}\approx -0-2.716341211i$$

$$x=-\frac{\sqrt{6\sqrt{31}-33}}{3}\approx -0.212547035$$

$$x=\frac{\sqrt{6\sqrt{31}-33}}{3}\approx 0.212547035$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+1.

$$\left(4x^{2}+4\right)\left(2x^{2}+1\right)=5\left(x^{2}-1\right)^{2}$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x^{2}+4 ki te 2x^{2}+1 ka whakakotahi i ngā kupu rite.

$$8x^{4}+12x^{2}+4=5\left(x^{2}-1\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x^{2}-1\right)^{2}.

$$8x^{4}+12x^{2}+4=5\left(\left(x^{2}\right)^{2}-2x^{2}+1\right)$$

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

$$8x^{4}+12x^{2}+4=5\left(x^{4}-2x^{2}+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5 ki te x^{4}-2x^{2}+1.

$$8x^{4}+12x^{2}+4=5x^{4}-10x^{2}+5$$

Tangohia te 5x^{4} mai i ngā taha e rua.

$$8x^{4}+12x^{2}+4-5x^{4}=-10x^{2}+5$$

Pahekotia te 8x^{4} me -5x^{4}, ka 3x^{4}.

$$3x^{4}+12x^{2}+4=-10x^{2}+5$$

Me tāpiri te 10x^{2} ki ngā taha e rua.

$$3x^{4}+12x^{2}+4+10x^{2}=5$$

Pahekotia te 12x^{2} me 10x^{2}, ka 22x^{2}.

$$3x^{4}+22x^{2}+4=5$$

Tangohia te 5 mai i ngā taha e rua.

$$3x^{4}+22x^{2}+4-5=0$$

Tangohia te 5 i te 4, ka -1.

$$3x^{4}+22x^{2}-1=0$$

Whakakapia te t mō te x^{2}.

$$3t^{2}+22t-1=0$$

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 3 mō te a, te 22 mō te b, me te -1 mō te c i te ture pūrua.

$$t=\frac{-22±\sqrt{22^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$t=\frac{-22±4\sqrt{31}}{6}$$

Whakaotia te whārite t=\frac{-22±4\sqrt{31}}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{2\sqrt{31}-11}{3} t=\frac{-2\sqrt{31}-11}{3}$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

$$x=-\sqrt{\frac{2\sqrt{31}-11}{3}} x=\sqrt{\frac{2\sqrt{31}-11}{3}} x=-i\sqrt{\frac{2\sqrt{31}+11}{3}} x=i\sqrt{\frac{2\sqrt{31}+11}{3}}$$

Whakaoti mō x

$$x=-\frac{\sqrt{6\sqrt{31}-33}}{3}\approx -0.212547035$$

$$x=\frac{\sqrt{6\sqrt{31}-33}}{3}\approx 0.212547035$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+1.

$$\left(4x^{2}+4\right)\left(2x^{2}+1\right)=5\left(x^{2}-1\right)^{2}$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x^{2}+4 ki te 2x^{2}+1 ka whakakotahi i ngā kupu rite.

$$8x^{4}+12x^{2}+4=5\left(x^{2}-1\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x^{2}-1\right)^{2}.

$$8x^{4}+12x^{2}+4=5\left(\left(x^{2}\right)^{2}-2x^{2}+1\right)$$

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

$$8x^{4}+12x^{2}+4=5\left(x^{4}-2x^{2}+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5 ki te x^{4}-2x^{2}+1.

$$8x^{4}+12x^{2}+4=5x^{4}-10x^{2}+5$$

Tangohia te 5x^{4} mai i ngā taha e rua.

$$8x^{4}+12x^{2}+4-5x^{4}=-10x^{2}+5$$

Pahekotia te 8x^{4} me -5x^{4}, ka 3x^{4}.

$$3x^{4}+12x^{2}+4=-10x^{2}+5$$

Me tāpiri te 10x^{2} ki ngā taha e rua.

$$3x^{4}+12x^{2}+4+10x^{2}=5$$

Pahekotia te 12x^{2} me 10x^{2}, ka 22x^{2}.

$$3x^{4}+22x^{2}+4=5$$

Tangohia te 5 mai i ngā taha e rua.

$$3x^{4}+22x^{2}+4-5=0$$

Tangohia te 5 i te 4, ka -1.

$$3x^{4}+22x^{2}-1=0$$

Whakakapia te t mō te x^{2}.

$$3t^{2}+22t-1=0$$

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 3 mō te a, te 22 mō te b, me te -1 mō te c i te ture pūrua.

$$t=\frac{-22±\sqrt{22^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Mahia ngā tātaitai.

$$t=\frac{-22±4\sqrt{31}}{6}$$

Whakaotia te whārite t=\frac{-22±4\sqrt{31}}{6} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{2\sqrt{31}-11}{3} t=\frac{-2\sqrt{31}-11}{3}$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō t tōrunga.

$$x=\sqrt{\frac{2\sqrt{31}-11}{3}} x=-\sqrt{\frac{2\sqrt{31}-11}{3}}$$