Whakaoti i te x^4-15x^2=16

$${ x }^{ 4 } -15 { x }^{ 2 } =16$$

Whakaoti mō x (complex solution)

$$x=4$$

$$x=-4$$

$$x=-i$$

$$x=i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 16 mai i ngā taha e rua.

$$x^{4}-15x^{2}-16=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}-15t-16=0$$

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 -15 mō te b, me te -16 mō te c i te ture pūrua.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 1\left(-16\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{15±17}{2}$$

Whakaotia te whārite t=\frac{15±17}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=16 t=-1$$

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

$$x=-4 x=4 x=-i x=i$$

Whakaoti mō x

$$x=-4$$

$$x=4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 16 mai i ngā taha e rua.

$$x^{4}-15x^{2}-16=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}-15t-16=0$$

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 -15 mō te b, me te -16 mō te c i te ture pūrua.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 1\left(-16\right)}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{15±17}{2}$$

Whakaotia te whārite t=\frac{15±17}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=16 t=-1$$

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

$$x=4 x=-4$$