Whakaoti i te -2/5(x^2+1/2)^2=-3/8

$$- \frac { 2 } { 5 } ( x ^ { 2 } + \frac { 1 } { 2 } ) ^ { 2 } = - \frac { 3 } { 8 }$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0.684284909$$

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0.684284909$$

$$x=-\frac{i\sqrt{\sqrt{15}+2}}{2}\approx -0-1.211711945i$$

$$x=\frac{i\sqrt{\sqrt{15}+2}}{2}\approx 1.211711945i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua ki te -\frac{5}{2}, te tau utu o -\frac{2}{5}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

Whakareatia te -\frac{3}{8} ki te -\frac{5}{2}, ka \frac{15}{16}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

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

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Tangohia te \frac{15}{16} mai i ngā taha e rua.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

Tangohia te \frac{15}{16} i te \frac{1}{4}, ka -\frac{11}{16}.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}+t-\frac{11}{16}=0$$

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 1 mō te b, me te -\frac{11}{16} mō te c i te ture pūrua.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

Whakaotia te whārite t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

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

Whakaoti mō x

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0.684284909$$

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0.684284909$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua ki te -\frac{5}{2}, te tau utu o -\frac{2}{5}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

Whakareatia te -\frac{3}{8} ki te -\frac{5}{2}, ka \frac{15}{16}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

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

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Tangohia te \frac{15}{16} mai i ngā taha e rua.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

Tangohia te \frac{15}{16} i te \frac{1}{4}, ka -\frac{11}{16}.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}+t-\frac{11}{16}=0$$

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 1 mō te b, me te -\frac{11}{16} mō te c i te ture pūrua.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

Whakaotia te whārite t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

$$x=\frac{\sqrt{\sqrt{15}-2}}{2} x=-\frac{\sqrt{\sqrt{15}-2}}{2}$$