Whakaoti i te (2x^2+2)^2-2(2x^2+2)-8=0

$$( 2 x ^ { 2 } + 2 ) ^ { 2 } - 2 ( 2 x ^ { 2 } + 2 ) - 8 = 0$$

Whakaoti mō x (complex solution)

$$x=1$$

$$x=-1$$

$$x=-\sqrt{2}i\approx -0-1.414213562i$$

$$x=\sqrt{2}i\approx 1.414213562i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$4\left(x^{2}\right)^{2}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-4x^{2}-4-8=0$$

Pahekotia te 8x^{2} me -4x^{2}, ka 4x^{2}.

$$4x^{4}+4x^{2}+4-4-8=0$$

Tangohia te 4 i te 4, ka 0.

$$4x^{4}+4x^{2}-8=0$$

Whakakapia te t mō te x^{2}.

$$4t^{2}+4t-8=0$$

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 -8 mō te c i te ture pūrua.

$$t=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$t=\frac{-4±12}{8}$$

Whakaotia te whārite t=\frac{-4±12}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$t=1 t=-2$$

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

$$x=-1 x=1 x=-\sqrt{2}i x=\sqrt{2}i$$

Whakaoti mō x

$$x=-1$$

$$x=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$4\left(x^{2}\right)^{2}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-4x^{2}-4-8=0$$

Pahekotia te 8x^{2} me -4x^{2}, ka 4x^{2}.

$$4x^{4}+4x^{2}+4-4-8=0$$

Tangohia te 4 i te 4, ka 0.

$$4x^{4}+4x^{2}-8=0$$

Whakakapia te t mō te x^{2}.

$$4t^{2}+4t-8=0$$

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 -8 mō te c i te ture pūrua.

$$t=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$t=\frac{-4±12}{8}$$

Whakaotia te whārite t=\frac{-4±12}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$t=1 t=-2$$

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

$$x=1 x=-1$$